Chào Thọ Nguyễn Đức. Nhằm mục tiêu phát triển bền vững. Từ tháng 082016, BQT 123doc.org kêu gọi thành viên chưa chia sẻ tài liệu trong năm vừa qua chia sẻ ít nhất 2 tài liệu miễn phínăm cho cộng đồng. Đây là lời kêu gọi dành cho các bạn biết cảm ơn những giá trị đã nhận được từ 123doc.org, và có trách nhiệm duy trì giá trị đó với cộng đồng. Bạn hãy UPLOAD chia sẻ ngay để trở thành một phần của lý tưởng tốt đẹp này
𝑇𝑢𝑦ể𝑛 𝑡ậ𝑝 𝑙ờ𝑖 𝑔𝑖ả𝑖 ℎ𝑎𝑦 𝑡ừ 𝑐á𝑐 𝑏à𝑖 ℎệ 1) { √1 + 𝑥 + √1 − 𝑦 = 2(1) 𝑥 − 𝑦 + 9𝑦 = 𝑥(9 + 𝑦 − 𝑦 )(2) 𝑝𝑡(2): 𝑥 − 𝑦 + 9𝑦 = 9𝑥 + 𝑥𝑦 − 𝑥𝑦 => 𝑥(𝑥 − 𝑦) + 𝑦 (𝑥 − 𝑦) − 9(𝑥 − 𝑦) = 𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 + 𝑦 − = 𝑣ớ𝑖 𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): √1 + 𝑥 + √1 − 𝑥 = 2(đ𝑘 − 𝑥 ≥ => 𝑥 ≤ 1) đặ𝑡 𝑡 = + 𝑥 => 𝑡 + √2 − 𝑡 = => − 𝑡 = − 4𝑡 + 𝑡 (đ𝑘 − 𝑡 ≥ => 𝑡 ≤ 2) => −𝑡 − 𝑡 + 4𝑡 − = => 𝑡 = ℎ𝑎𝑦 − 𝑡 − 2𝑡 + = 𝑡 = => 𝑥 = 𝑦 = 𝑡 = −1 + √3 => 𝑥 = 𝑦 = −11 + 6√3 𝑡 = −1 − √3 => 𝑥 = 𝑦 = −11 − 6√3 𝑣ớ𝑖 𝑥 = − 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): √10 − 𝑦 + √1 − 𝑦 = 2(𝑝𝑡𝑣𝑛) đặ𝑡 𝑓(𝑦) = √10 − 𝑦 + √1 − 𝑦 − 𝑓 ′ (𝑦) = −3𝑦 −1 √(10−𝑦 )2 + 2√1−𝑦 > 0(đ𝑘 𝑦 < 1) ℎà𝑚 𝑛à𝑦 𝑙𝑢ơ𝑛 đồ𝑛𝑔 𝑏𝑖ế𝑛 𝑣ậ𝑦 𝑝ℎả𝑖 𝑐ó í𝑡 𝑛ℎấ𝑡 𝑛𝑔ℎ𝑖ệ𝑚 𝑓(𝑦) = 𝑣ậ𝑦 𝑝𝑡 𝑛à𝑦 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 + √𝑥 2) {𝑥 𝑦2 𝑦 +1+𝑥 + 𝑦 = 0(1) + 2√𝑥 + + 𝑦 = 3(2) 𝑝𝑡(1): 𝑥 + 𝑦(√𝑥 + − 𝑥) + 𝑦 = => 𝑦√𝑥 + − 𝑥𝑦 = −𝑥 − 𝑦 => 𝑦√𝑥 + = 𝑥𝑦 − 𝑥 − 𝑦 2(𝑥𝑦−𝑥−𝑦 ) 𝑥2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 𝑦 + 𝑥 𝑦 + 𝑦2 = 𝑥 => (𝑦 + 𝑦) − (𝑦 + 𝑦) − = 𝑥 => 𝑦 + 𝑦 = ℎ𝑎𝑦 𝑥 𝑦 + 𝑦 = −1 𝑥 𝑣ớ𝑖 𝑦 + 𝑦 = => 𝑥 = 3𝑦 − 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑦√(3𝑦 − 𝑦 )2 + = (3𝑦 − 𝑦 )𝑦 − (3𝑦 − 𝑦 ) − 𝑦 => 𝑦√9𝑦 − 6𝑦 + 𝑦 + = 3𝑦 − 𝑦 − 3𝑦 + 𝑦 − 𝑦 => 𝑦√𝑦 − 6𝑦 + 9𝑦 + = −𝑦 + 3𝑦 − 3𝑦 𝑦 = 0(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 √𝑦 − 6𝑦 + 9𝑦 + = 3𝑦 − 𝑦 − 𝑉ớ𝑖 √𝑦 − 6𝑦 + 9𝑦 + = 3𝑦 − 𝑦 − => 𝑦 − 6𝑦 + 9𝑦 + = (3𝑦 − 𝑦 − 3)2 (đ𝑘 − 𝑦 + 3𝑦 − ≥ 0) => 𝑦 − 6𝑦 + 9𝑦 + = 9𝑦 + 𝑦 + − 6𝑦 + 6𝑦 − 18𝑦 => −6𝑦 + = − 6𝑦 + 6𝑦 − 18𝑦 => 6𝑦 − 18𝑦 + = => 𝑦 = 9+√33 (𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑦 = 9−√33 (𝑙𝑜ạ𝑖) 𝑥 𝑣ớ𝑖 𝑦 + 𝑦 = −1 => 𝑥 = −𝑦 − 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑦√(−𝑦 − 𝑦 )2 + = (−𝑦 − 𝑦 )𝑦 − (−𝑦 − 𝑦 ) − 𝑦 => 𝑦√𝑦 + 2𝑦 + 𝑦 + = −𝑦 − 𝑦 + 𝑦 + 𝑦 − 𝑦 => 𝑦√𝑦 + 2𝑦 + 𝑦 + = −𝑦 − 𝑦 + 𝑦 => √𝑦 + 2𝑦 + 𝑦 + = − 𝑦 − 𝑦 => 𝑦 + 2𝑦 + 𝑦 + = (1 − 𝑦 − 𝑦)2 (đ𝑘: − 𝑦 − 𝑦 ≥ 0) => 𝑦 + 2𝑦 + 𝑦 + = + 𝑦 + 𝑦 − 2𝑦 + 2𝑦 − 2𝑦 => 𝑦 + 2𝑦 + 𝑦 + = 𝑦 + 2𝑦 − 𝑦 − 2𝑦 + => 2𝑦 + 2𝑦 = => 𝑦 = −1(𝑡ℎõ𝑎) 𝑦 = −1 => 𝑥 = 3) { 𝑥(𝑥𝑦 + 𝑥)2 + (𝑥 + 1)2 = 𝑥 (𝑦 + 𝑦 + 1) + 𝑥 (𝑦 − 1) + 5𝑥(1) 4𝑥 𝑦 + 7𝑥 + 2𝑥 √𝑦 + = 2𝑥 + 1(2) 𝑝𝑡(1): 𝑥(𝑥𝑦 + 𝑥)2 + (𝑥 + 1)2 = 𝑥 (𝑦 + 𝑦 + 1) + 𝑥 (𝑦 − 1) + 5𝑥 => 𝑥(𝑥 𝑦 + 2𝑥 𝑦 + 𝑥 ) + (𝑥 + 2𝑥 + 1) = 𝑥 (𝑦 + 𝑦 + 1) + 𝑥 (𝑦 − 1) + 5𝑥 => 𝑥 𝑦 + 2𝑥 𝑦 + 𝑥 + 𝑥 + 2𝑥 + = 𝑥 𝑦 + 𝑥 𝑦 + 𝑥 + 𝑥 𝑦 − 𝑥 + 5𝑥 => 𝑥 𝑦 + 2𝑥 − 3𝑥 + = 𝑥 𝑦 => 𝑥 𝑦(𝑥 − 1) + (𝑥 − 1)(2𝑥 − 1) = 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑥 𝑦 + 2𝑥 − = 𝑣ớ𝑖 𝑥 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 4𝑥 𝑦 + 7𝑥 + 2𝑥 √𝑦 + = 2𝑥 + => 4𝑦 + + 2√𝑦 + = => 𝑦 = −1 ℎ𝑎𝑦 4√𝑦 + + = 0(𝑝𝑡𝑣𝑛) 𝑣ớ𝑖 𝑥 𝑦 + 2𝑥 − = => 𝑦 = 1−2𝑥 𝑥2 (𝑥é𝑡 𝑥 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎệ) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 4𝑥 ( 1−2𝑥 𝑥2 1−2𝑥 ) + 7𝑥 + 2𝑥 √ 𝑥2 + = 2𝑥 + 1−2𝑥+𝑥 => 4𝑥(1 − 2𝑥) + 7𝑥 − 2𝑥 − + 2𝑥 √ 𝑥2 =0 => −𝑥 + 2𝑥 − + 2𝑥√(1 − 𝑥)2 = => −𝑥 + 2𝑥 − + 2𝑥(1 − 𝑥) = => −𝑥 + 2𝑥 − + 2𝑥 − 2𝑥 = => 𝑥 = ℎ𝑎𝑦 𝑥 = 𝑥 = => 𝑦 = 𝑥 (21𝑦 − 20) − = 0(1) 4) { 𝑥(𝑦 + 20) − 21 = 0(2) 𝑡𝑎 𝑑ễ 𝑑à𝑛𝑔 𝑛ℎậ𝑛 𝑡ℎấ𝑦 𝑥 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎệ => { 21𝑦 − 20 − 𝑥 = 0(1) 𝑦 + 20 − 21 𝑥 = 0(2) 𝑐ộ𝑛𝑔 𝑝𝑡(1), 𝑝𝑡(2): 21𝑦 + 𝑦 = 𝑥 + 21 𝑥 𝑥é𝑡 𝑓(𝑡) = 𝑡 + 21𝑡 => 𝑓 ′ (𝑡) = 3𝑡 + 21 > 1 𝑣ậ𝑦 𝑦 = 𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 21𝑦 − 20 − 𝑥 = => 21𝑦 − 20 − 𝑦 = => 𝑦 = −5 ℎ𝑎𝑦 𝑦 = ℎ𝑎𝑦 𝑦 = 1 𝑦 = −5 => 𝑥 = − 𝑦 = => 𝑥 = 𝑦 = => 𝑥 = 𝑥(𝑥 + 𝑦) + √𝑥 + 𝑦 = √2𝑦(√2𝑦 + 1)(1) 5) { 8𝑥 − 8𝑦 + 2(𝑥 − 𝑦 ) + = 8𝑦√2𝑥 − 3𝑥 + 1(2) 𝑝𝑡(1): 𝑥(𝑥 + 𝑦) + √𝑥 + 𝑦 = √2𝑦(√2𝑦 + 1) (đ𝑘: { 𝑦≥0 ) 𝑥+𝑦 ≥0 => 𝑥 + 𝑥𝑦 + √𝑥 + 𝑦 = 2𝑦 + √2𝑦 => 𝑥 + 𝑥𝑦 − 2𝑦 + 𝑥−𝑦 √𝑥+𝑦+√2𝑦 => (𝑥 − 𝑦)(𝑥 + 2𝑦) + =0 𝑥−𝑦 √𝑥+𝑦+√2𝑦 𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 + 2𝑦 + =0 √𝑥+𝑦+√2𝑦 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0) 𝑣ớ𝑖 𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 8𝑥 − 8𝑦 + 2(𝑥 − 𝑦 ) + = 8𝑦√2𝑥 − 3𝑥 + => 8𝑥 − 8𝑥 + = 8𝑥√2𝑥 − 3𝑥 + 1(đ𝑘: 2𝑥 − 3𝑥 + ≥ 0) => 8𝑥 − 11𝑥 + = 8𝑥√2𝑥 − 3𝑥 + − => 8𝑥 − 11𝑥 + = 8𝑥 √2𝑥 −3𝑥+1+ 8𝑥(2𝑥 −3𝑥+ ) 𝑣ậ𝑦 8𝑥 − 11𝑥 + = ℎ𝑎𝑦 = 2𝑥 √2𝑥 −3𝑥+1+ 8𝑥 − 11𝑥 + = => 𝑥 = 𝑦 = ℎ𝑎𝑦 𝑥 = 𝑦 = 1 √2𝑥 − 3𝑥 + + = 2𝑥 => √2𝑥 − 3𝑥 + = 2𝑥 − (đ𝑘 𝑥 ≥ 4) => 2𝑥 − 3𝑥 + = 4𝑥 − 2𝑥 + => 𝑥 = 𝑦 = −1+√7 ℎ𝑎𝑦 𝑥 = −1−√7 (𝑙𝑜ạ𝑖) 𝑥 − 2𝑥 = 𝑦 − 𝑦(1) 6) { (𝑥 − 𝑦 )3 = 3(2) đặ𝑡 𝑎 = 𝑥 + 𝑦, 𝑏 = 𝑥 − 𝑦 => 𝑎 + 𝑏 = 2𝑥 𝑣à 𝑎 − 𝑏 = 2𝑦 𝑝𝑡(1): (𝑥 − 𝑦 )(𝑥 + 𝑦 ) = 2𝑥 − 𝑦 => 𝑎𝑏 (( 𝑎+𝑏 2 ) +( 𝑎2 +𝑏2 => 𝑎𝑏 ( )= 𝑎−𝑏 ) ) = (𝑎 + 𝑏) − 𝑎−𝑏 𝑎+3𝑏 𝑝𝑡(2): 𝑎3 𝑏 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑎𝑏(𝑎2 + 𝑏 ) = 𝑎 + 𝑎3 𝑏 => 𝑎3 𝑏 + 𝑎𝑏 = 𝑎 + 𝑎3 𝑏 => 𝑎(𝑎2 𝑏 − 1) = 𝑎𝑏 (𝑎2 𝑏 − 1) => (𝑎2 𝑏 − 1)(𝑏 − 1)𝑎 = 𝑎 = ℎ𝑎𝑦 𝑎2 𝑏 − = ℎ𝑎𝑦 𝑏 = 𝑎 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎệ 𝑣ớ𝑖 𝑎2 𝑏 = => 𝑏 = 𝑎2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 𝑎3 = => 𝑎 = √3 => 𝑏 = √9 1 3 + √9 √3 𝑥+𝑦= 𝑥= √3 => { 𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: { 𝑥 − 𝑦 = √9 𝑦 =−3 √3 𝑣ớ𝑖 𝑏 = => 𝑎 = √3 𝑥= 𝑥 + 𝑦 = √3 𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: { => { 𝑥−𝑦 =1 𝑦= √3+1 √3−1 𝑥 + 𝑦 + 𝑥𝑦 = 4𝑦 − 1(1) 7) { 𝑦 𝑥 + 𝑦 − = 1+𝑥 (2) 𝑝𝑡(1): (𝑥 + 1) + 𝑦 + 𝑥𝑦 − 4𝑦 = => (𝑥 + 1) + 𝑦(𝑥 + 𝑦 − 4) = 𝑥é𝑡 𝑦 = => 𝑥 = −1(𝑝𝑡𝑣𝑛) 𝑥 +1 𝑦 ≠ => 𝑦 +𝑥+𝑦−4=0 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑚ớ𝑖: { 𝑥+𝑦−4+ 𝑥 +1 𝑦 𝑦 = 0(1) 𝑥 + 𝑦 − − 1+𝑥 = 0(2) đặ𝑡 𝑎 = { 𝑥 +1 𝑦 ,𝑏 = 𝑥 + 𝑦 𝑎+𝑏−4=0 𝑏 =4−𝑎 => {4 − 𝑎 − − = 𝑏−𝑎−2=0 𝑎 => − 𝑎 − 𝑎 = => 2𝑎 − 𝑎2 − = => −𝑎2 + 2𝑎 − = => 𝑎 = => 𝑏 = 𝑥 +1 => { 𝑦 =1 => { 𝑥+𝑦 =3 (3 − 𝑦)2 + = 𝑦 𝑥 =3−𝑦 − 6𝑦 + 𝑦 + = 𝑦 => 𝑦 = ℎ𝑎𝑦 𝑦 = 𝑦 = => 𝑥 = −2 𝑦 = => 𝑥 = 𝑥 − 2𝑥𝑦 + 𝑥 + 𝑦 = 0(1) 8) { 𝑥 − 4𝑥 𝑦 + 3𝑥 + 𝑦 = 0(2) 𝑡𝑎 𝑑ễ 𝑑à𝑛𝑔 𝑡ℎấ𝑦 đượ𝑐 𝑥 = 𝑦 = 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑝𝑡(1): 𝑥 + 𝑦(1 − 2𝑥) + 𝑥 = 𝑥+𝑥 𝑦 = 2𝑥−1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 𝑥 − 4𝑥 (𝑥+𝑥 ) 𝑥+𝑥 2 + 3𝑥 + (2𝑥−1) = 2𝑥−1 => 𝑥 (2𝑥 − 1)2 − 4𝑥 (𝑥 + 𝑥)(2𝑥 − 1) + 3𝑥 (2𝑥 − 1)2 + (𝑥 + 𝑥 )2 = => 2𝑥 (𝑥 − 2)(𝑥 − 1)(2𝑥 + 1) = 𝑥 = ℎ𝑎𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = 𝑥 = => 𝑦 = 𝑥 = => 𝑦 = 𝑥 − 2𝑥𝑦 + 𝑥 + 𝑦 = 0(1) 𝐶á𝑐ℎ 2: { 𝑥 − 4𝑥 𝑦 + 3𝑥 + 𝑦 = 0(2) => { 𝑥 + 𝑦 = 2𝑥𝑦 − 𝑥(1) (𝑥 + 𝑦)2 − 6𝑥 𝑦 + 3𝑥 = 0(2) 𝑙ấ𝑦 𝑝𝑡(1)𝑡ℎế 𝑝𝑡(2): (2𝑥𝑦 − 𝑥)2 − 6𝑥 𝑦 + 3𝑥 = => 𝑥 (2𝑦 − 1)2 + 3𝑥 (1 − 2𝑦) = => 𝑥 = ℎ𝑎𝑦 (2𝑦 − 1)2 − 3(2𝑦 − 1) = => 𝑦 = ℎ𝑎𝑦 𝑦 = 𝑥 = => 𝑦 = 𝑦 = => 𝑥 ∈ ∅ 𝑦 = => 𝑥 = ℎ𝑎𝑦 𝑥 = 𝑥 + 2𝑥 𝑦 + 𝑥 𝑦 = 2𝑥 + 9(1) 9) { 𝑥 + 2𝑥𝑦 = 6𝑥 + 6(2) 𝑝𝑡(1): (𝑥 + 𝑥𝑦)2 = 2𝑥 + 𝑝𝑡(2): 𝑥 + 𝑥𝑦 = 6𝑥 + − 𝑥𝑦 𝑙ấ𝑦 𝑝𝑡(2)𝑡ℎế 𝑝𝑡(1): (6𝑥 + − 𝑥𝑦)2 = 2𝑥 + 9(3) 𝑝𝑡(2): 𝑦 = 6𝑥+6−𝑥 2𝑥 => (6𝑥 + − => (6𝑥 + − (𝑥 ≠ 0) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(3): 𝑥(6𝑥+6−𝑥 ) 2𝑥 6𝑥+6−𝑥 2 ) = 2𝑥 + ) = 2𝑥 + => 𝑥4 + 3𝑥 + 12𝑥 + 16𝑥 = => 𝑥 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑥 = −4 𝑥 = −4 => 𝑦 = 17 2𝑥−5𝑦+4 10) { √𝑥 − 𝑦 − = 6𝑦−4𝑥+6 (1) 4𝑥 − (12𝑦 − 5)𝑥 + 9𝑦 − 4𝑦 − 11 = (8𝑥 − 12𝑦)√𝑥 − 𝑦 − 1(2) 𝑝𝑡(1): đặ𝑡 𝑡 = 𝑥 − 𝑦 − => 𝜕𝑡 − 𝑡(6𝑦 − 4𝑥 + 6) + (2𝑥 − 5𝑦 + 4) − 𝜕(𝑥 − 𝑦 − 1) = 𝑃 = √𝑑𝑒𝑛𝑡𝑎 = √(6𝑦 − 4𝑥 + 6)2 − 4𝜕[(2𝑥 − 5𝑦 + 4) − 𝜕(𝑥 − 𝑦 − 1)] 𝑐ℎ𝑜 𝑥 = 100, 𝑦 = 1000 𝑣à 𝑐ℎọ𝑛 𝑠ố 𝜕 𝑠𝑎𝑜 𝑐ℎ𝑜 𝑃 𝑙à 𝑠ố 𝑛𝑔𝑢𝑦ê𝑛 𝑘ℎ𝑖 𝑛à𝑦 𝜕 = −4 𝑙à 𝑠ố 𝑝ℎù ℎợ𝑝 => −4𝑡 + 𝑡(6𝑦 − 4𝑥 + 6) + (2𝑥 − 5𝑦 + 4) + 4(𝑥 − 𝑦 − 1) = 𝑑𝑒𝑛𝑡𝑎 = (6𝑦 − 4𝑥 + 6)2 + 16(6𝑥 − 9𝑦) = (4𝑥 − 6𝑦 + 6)2 ≥ 2𝑥−3𝑦 √𝑥−𝑦−1=−2(𝑝𝑡𝑣𝑛) => [ √𝑥−𝑦−1= 𝑣ớ𝑖 √𝑥 − 𝑦 − = 2𝑥−3𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => (2𝑥 − 3𝑦)2 − 2(2𝑥 − 3𝑦)(2√𝑥 − 𝑦 − 1) + 4(𝑥 − 𝑦 − 1) + 𝑥 − = => (2𝑥 − 3𝑦 − 2√𝑥 − 𝑦 − 1) + 𝑥 = => 𝑥 = 𝑡ℎế 𝑣à𝑜 𝑡ì𝑚 𝑦 => 2√6 − 𝑦 = 14 − 3𝑦 => 𝑦 = 11) { 40−2√13 𝑥 + 3𝑥𝑦 + 49 = 0(1) 𝑥 − 8𝑥𝑦 + 𝑦 = 8𝑦 − 17𝑥(2) 𝑡𝑎 đ𝑜á𝑛 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = −1 −3𝑦 + 48 = −3(𝑦 − 16) = { => { 𝑦 − 16 = 𝑦 − 16 = 𝑣ậ𝑦 − 3𝑝𝑡(2) − 𝑝𝑡(1): −3(𝑥 − 8𝑥𝑦 + 𝑦 − 8𝑦 + 17𝑥) − (𝑥 + 3𝑥𝑦 + 49) = => −𝑥 − 3𝑥 − 3𝑥𝑦 + 24𝑥𝑦 − 51𝑥 − 3𝑦 + 24𝑦 − 49 = => −(𝑥 + 1)(𝑥 + 2𝑥 + 3𝑦 − 24𝑦 + 49) = 𝑣ậ𝑦 𝑥 = −1 ℎ𝑎𝑦 𝑥 + 2𝑥 + 3𝑦 − 24𝑦 + 49 = 𝑥 = −1 => 𝑦 = ℎ𝑎𝑦 𝑦 = −4 { 𝑥 + 3𝑦 + 2𝑥 − 24𝑦 + 49 = 0(3) 𝑥 − 8𝑥𝑦 + 𝑦 − 8𝑦 + 17𝑥 = 0(2) 𝑙ấ𝑦 𝑝𝑡(3) − 𝑝𝑡(2): 3𝑦 + 2𝑥 − 24𝑦 + 49 + 8𝑥𝑦 − 𝑦 + 8𝑦 − 17𝑥 = => 2𝑦 − 15𝑥 − 16𝑦 + 49 + 8𝑥𝑦 = => 2𝑦 − 16𝑦 + 49 + 𝑥(−15 + 8𝑦) = => 𝑥 = 16𝑦−49−2𝑦 −15+8𝑦 16𝑦−49−2𝑦 => ( 8𝑦−15 (𝑦 ≠ 15 ) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(3): 16𝑦−49−2𝑦2 ) + 3𝑦 + ( 8𝑦−15 ) − 24𝑦 + 49 = => 49(𝑦 − 4)2 (4𝑦 − 16𝑦 + 19) = => 𝑦 = => 𝑥 = −1 𝑥 + 2𝑥 − 5𝑥 + 𝑦 − 6𝑥 − 11 = 0(1) 12) { 3√𝑦 −7−6 𝑥2 + 𝑥 = (2) √𝑦 −7 𝑝𝑡(1): 𝑦 − = −𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 𝑥 + 𝑥 = 3√−𝑥 −2𝑥 +5𝑥 +6𝑥+4−6 √−𝑥4 −2𝑥 +5𝑥 +6𝑥+4 3(−𝑥 −2𝑥 +5𝑥 +6𝑥) => 𝑥(𝑥 + 1)√−𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + = √−𝑥 −2𝑥 +5𝑥2 +6𝑥+4+2 −3𝑥(𝑥+1)(𝑥−2)(𝑥+3) => 𝑥(𝑥 + 1)√−𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + = √−𝑥 −2𝑥 +5𝑥2 +6𝑥+4+2 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = −1 ℎ𝑎𝑦 √– 𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + = √−𝑥4 −3(𝑥−2)(𝑥+3) −2𝑥 +5𝑥 +6𝑥+4+2 => −𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + + 3(𝑥 − 2)(𝑥 + 3) + 2√−𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + = => −(𝑥 + 𝑥)2 + 9(𝑥 + 𝑥) − + 2√−𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + − = −2(𝑥 +𝑥−5)(𝑥 +𝑥−1) => −(𝑥 + 𝑥 − 8)(𝑥 + 𝑥 − 1) + √−𝑥 −2𝑥 +5𝑥 +6𝑥+4+3 𝑣ậ𝑦 𝑥 + 𝑥 − = ℎ𝑎𝑦 − (𝑥 + 𝑥 − 8) + √−𝑥 =0 −2(𝑥 +𝑥−5) −2𝑥3 +5𝑥 +6𝑥+4+3 =0 𝑝𝑡 𝑐ò𝑛 𝑙ạ𝑖 đ𝑘: −𝑥 − 2𝑥 + 5𝑥 + 6𝑥 + ≥ 𝑥∈[ −1−√13+4√13 √13+4√13−1 , ] => 𝑥 + 𝑥 − > 𝑣ậ𝑦 𝑉𝑇 < => 𝑝𝑡 𝑡𝑟ê𝑛 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 13) { 7𝑦 = 4𝑥 + 3𝑥𝑦 (1) 7𝑥 = 𝑦 + 6𝑥 𝑦(2) 𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): 7(𝑥 + 𝑦) = 4𝑥 + 6𝑥 𝑦 + 3𝑥𝑦 + 𝑦 = (𝑥 + 𝑦)(4𝑥 + 2𝑥𝑦 + 𝑦 ) 𝑣ậ𝑦 𝑥 + 𝑦 = ℎ𝑎𝑦 4𝑥 + 2𝑥𝑦 + 𝑦 = 𝑣ớ𝑖 𝑥 = −𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 7𝑦 = 4𝑥 + 3𝑥𝑦 => −7𝑦 = 7𝑦 => 𝑦 = => 𝑥 = ℎ𝑎𝑦 𝑦 = −1(𝑝𝑡𝑣𝑛) 𝑣ớ𝑖 4𝑥 + 2𝑥𝑦 + 𝑦 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 4𝑥 + 3𝑥𝑦 = (4𝑥 + 2𝑥𝑦 + 𝑦 )𝑦 => 𝑦 = 𝑥 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 7𝑥 = 7𝑥 => 𝑥 = ℎ𝑎𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = −1 𝑥 + 𝑦 = 2(1) 14) { 2𝑥 𝑥+𝑦 + (𝑥𝑦 + 1)2 = 5(2) 𝑝𝑡(2): 2𝑥 + (𝑥𝑦 + 1)2 (𝑥 + 𝑦) = 5(𝑥 + 𝑦) => 2𝑥 + (𝑥𝑦 + 1)2 (𝑥 + 𝑦) = (𝑥 + 𝑦 )2 (𝑥 + 𝑦) + 𝑥 + 𝑦 => 2𝑥 + 𝑥𝑦(𝑥𝑦 + 𝑥 + 𝑦 )(𝑥 + 𝑦) = (𝑥 + 𝑦 )2 (𝑥 + 𝑦)(𝑡ℎế 𝑡ừ 𝑝𝑡(1)𝑣à𝑜) => 𝑥 = 𝑦 => 𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑥 = ℎ𝑎𝑦 𝑥 = −1 15) { 𝑥 + 𝑦 + 𝑥 + 𝑦 = 5(1) (𝑥𝑦 − 1)2 = 𝑥 − 𝑦 + 2(2) 1 𝑝𝑡(1): 𝑥 + 𝑦 + 𝑥 + 𝑦 = 𝑣ớ𝑖 𝑥 + − √2(𝑥 + 1) = => 𝑥 = 1±√2√2−1 √2 𝑣ớ𝑖 𝑥 + + √2(𝑥 + 1) = => 𝑝𝑡𝑣𝑛 5√+𝑥 81) √1−𝑥+𝑥 +2√1+𝑥 = 4𝑥 − 5𝑥 + đặ𝑡 𝑎 = √1 + 𝑥, 𝑏 = √1 − 𝑥 + 𝑥 (đ𝑘: 𝑎, 𝑏 ≥ 0) => 𝑎2 = + 𝑥, 4𝑏 = − 4𝑥 + 4𝑥 𝑎2 + 4𝑏 + 3𝑥 = 4𝑥 + 𝑡ℎế 𝑛𝑔ượ𝑐 𝑣à𝑜 𝑝𝑡 𝑡𝑟ê𝑛 𝑡𝑎 đượ𝑐 5𝑎 => 𝑏+2𝑎 = 𝑎2 + 4𝑏 − 2(𝑎2 − 1) = 4𝑏 − 𝑎2 + => 5𝑎 = (𝑏 + 2𝑎)(4𝑏 − 𝑎2 + 2) => 2𝑎3 + 𝑎2 𝑏 − 8𝑎𝑏 − 4𝑏 + (𝑎 − 2𝑏) = => (𝑎 − 2𝑏)(2𝑎 + 𝑏)(𝑎 + 2𝑏) + (𝑎 − 2𝑏) = 𝑣ậ𝑦 𝑎 − 2𝑏 = ℎ𝑎𝑦 (2𝑎 + 𝑏)(𝑎 + 2𝑏) + = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 ≥ 1) 𝑣ớ𝑖 𝑎 − 2𝑏 = => √1 + 𝑥 = 2√1 − 𝑥 + 𝑥 (𝑝𝑡𝑣𝑛) 82) (𝑥 + 2)√𝑥 + 𝑥 + + 𝑥 − 3𝑥 − 5𝑥 + = đặ𝑡 𝑡 = √𝑥 + 𝑥 + 𝛼𝑡 + (𝑥 + 2)𝑡 + 𝑥 − 3𝑥 − 5𝑥 + − 𝑎(𝑥 + 𝑥 + 1) = 𝑃 = √𝑑𝑒𝑛𝑡𝑎 = √(𝑥 + 2)2 − 4𝛼(𝑥 − 3𝑥 − 5𝑥 + − 𝛼(𝑥 + 𝑥 + 1)) 𝑔á𝑛 𝑥 = 100 𝑡ì𝑚 𝛼 𝑠𝑎𝑜 𝑐ℎ𝑜 𝑃 𝑙à 𝑠ố 𝑛𝑔𝑢𝑦ê𝑛 𝑣ậ𝑦 𝛼 = −1 𝑡ℎì 𝑡ℎõ𝑎 𝑚ã𝑛 −𝑡 + (𝑥 + 2)𝑡 + 𝑥 − 2𝑥 − 4𝑥 + 𝑑𝑒𝑛𝑡𝑎 = (𝑥 + 2)2 + 4(𝑥 − 2𝑥 − 4𝑥 + 3) = (𝑥 + 2𝑥 − 4)2 ≥ 𝑣ậ𝑦 √𝑥 + 𝑥 + = 𝑥 + 𝑥 − ℎ𝑎𝑦 √𝑥 + 𝑥 + = − 𝑥 𝑣ớ𝑖 √𝑥 + 𝑥 + = 𝑥 + 𝑥 − => 𝑥 + 𝑥 + = (𝑥 + 𝑥 − 1)2 (đ𝑘: 𝑥 + 𝑥 − ≥ 0) => 𝑥 + 𝑥 + = 𝑥 + 𝑥 + + 2𝑥 − 2𝑥 − 2𝑥 => 𝑥 + 𝑥 + = 𝑥 − 𝑥 + 2𝑥 − 2𝑥 + => 𝑥 + 2𝑥 − 2𝑥 − 3𝑥 = => 𝑥 = 0(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑥 = −1(𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑥 = − 1±√13 𝑣ớ𝑖 √𝑥 + 𝑥 + = − 𝑥 => 𝑥 + 𝑥 + = (3 − 𝑥)2 (đ𝑘: − 𝑥 ≥ => 𝑥 ≤ 3) => 𝑥 = (𝑛ℎậ𝑛) 83) { (𝑥 + 1)2 + (𝑥 + 1)√𝑦 + + 𝑦 = 6(1) 𝑥 + (2 + 𝑥)√𝑦 + = 4(2) để ý 𝑝𝑡(1): (𝑥 + 1)2 + (𝑥 + 1)√𝑦 + + (𝑦 + 1) = 𝑝𝑡(2): (𝑥 + 1) + (𝑥 + 1)√𝑦 + + √𝑦 + = 𝑐ộ𝑛𝑔 𝑝𝑡(1)𝑣à 𝑝𝑡(2)𝑙ạ𝑖: (𝑥 + 1)2 + 2(𝑥 + 1)√𝑦 + + (𝑦 + 1) + (𝑥 + 1) + √𝑦 + = 12 => (𝑥 + + √𝑦 + 1) + (𝑥 + + √𝑦 + 1) − 12 = 𝑣ậ𝑦 𝑥 + + √𝑦 + = ℎ𝑎𝑦 𝑥 + + √𝑦 + = −4 𝑣ớ𝑖 𝑥 + + √𝑦 + = 𝑘ế𝑡 ℎợ𝑝 𝑣ớ𝑖 𝑝𝑡(1) 𝑡𝑎 đượ𝑐 𝑥 = ℎ𝑎𝑦 𝑥 = 𝑣ớ𝑖 𝑥 + + √𝑦 + = −4 𝑘ế𝑡 ℎợ𝑝 𝑣ớ𝑖 𝑝𝑡(2) 𝑝𝑡𝑣𝑛 84) { (𝑥 + 3𝑦 + 1)√2𝑥𝑦 + 2𝑦 = 𝑦(3𝑥 + 4𝑦 + 3)(1) (√𝑥 + − √2𝑦 − 2)(𝑥 − + √𝑥 + 𝑥 + 2𝑦 − 4) = 4(2) để ý 𝑝𝑡(1): (𝑥 + 1)√2𝑦(𝑥 + 1) + 3𝑦√2𝑦(𝑥 + 1) = 3(𝑥 + 1)𝑦 + 4𝑦 đặ𝑡 𝑎2 = 𝑥 + 1, 𝑏 = 2𝑦(đ𝑘: 𝑎, 𝑏 ≥ 0) 3 3 => (√𝑥 + 1) √2𝑦 + (√2𝑦) √𝑥 + = (𝑥 + 1)2𝑦 + (2𝑦)2 => 𝑎3 𝑏 + 3𝑏 𝑎 = 3𝑎2 𝑏 2 + 𝑏4 => 𝑏(𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 ) + 3𝑎𝑏 2 𝑣ậ𝑦 𝑎 = 𝑏 ℎ𝑎𝑦 𝑏(𝑎2 + 𝑎𝑏 + 𝑏 ) − (𝑏 − 𝑎) = 3𝑎𝑏 𝑣ớ𝑖 𝑝𝑡(∗): 𝑏(2𝑎2 − 𝑎𝑏 + 2𝑏 ) = = 0(∗) 𝑣ậ𝑦 𝑏 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 2𝑎2 − 𝑎𝑏 + 2𝑏 = 𝑣ớ𝑖 2𝑎2 − 𝑎𝑏 + 2𝑏 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑎, 𝑏 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎệ) 𝑣ớ𝑖 𝑎 = 𝑏 => √𝑥 + = √2𝑦 => 𝑦 = 𝑥+1 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => (√𝑥 + − √𝑥 − 1)(𝑥 − + √𝑥 + 2𝑥 − 3) = => 𝑥 − + √𝑥 + 2𝑥 − = √𝑥 + + √𝑥 − 𝑔ọ𝑖 𝑡 = √𝑥 + 3, 𝑘 = √𝑥 − 1(đ𝑘: 𝑡, 𝑘 ≥ 0) 𝑚+𝑛 =1 đă𝑡 𝑚(𝑥 + 3) + 𝑛(𝑥 − 1) = 𝑥 − => { 3𝑚 − 𝑛 = −3 => − 𝑡2 + 𝑡𝑘 + 3𝑘 2 =𝑡+𝑘 => − (𝑡 − 3𝑘)(𝑡 + 𝑘) = 𝑡 + 𝑘 => 3𝑘 − 𝑡 = => 3√𝑥 − − √𝑥 + = => 𝑥 = 85) √𝑥 + 5𝑥 − = √ 13 5𝑥 −2 5𝑥 −2 đặ𝑡 𝑡 = 𝑥 + 5𝑥 (∗∗) => 𝑡 − = √ => 6(𝑡 − 2𝑡 + 1) = 5𝑥 − 2(đ𝑘: 𝑡 ≥ 1) => 6𝑡 − 12𝑡 + = 5𝑥 (∗) 𝑙ấ𝑦 𝑝𝑡(∗) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗∗): 𝑡 = 𝑥 + 6𝑡 − 12𝑡 + => 𝑡 − 6𝑡 + 12𝑡 − = 𝑥 3 => (𝑡 − 2)3 = 𝑥 => √𝑥 + 5𝑥 − = 𝑥 => 𝑥 = −6 ± 2√7 𝑥 + 𝑦 = 9(1) 86) { 1 1 ( + 𝑦) (1 + ) (1 + 𝑦) = 18(2) √𝑥 đặ𝑡 𝑎 = √𝑥 √ √𝑥 ,𝑏 = √𝑦 √ 𝑎3 + 𝑏 = 9(1) 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑚ớ𝑖: { (𝑎 + 𝑏)(1 + 𝑎)(1 + 𝑏) = 18(2) 𝑝𝑡(2): 𝑎2 𝑏 + 𝑎𝑏 + (𝑎 + 𝑏)2 + (𝑎 + 𝑏) = 18 𝑛ℎâ𝑛 3𝑝𝑡(2) + 𝑝𝑡(1): (𝑎 + 𝑏)3 + 3(𝑎 + 𝑏)2 + 3(𝑎 + 𝑏) = 63 => (𝑎 + 𝑏 + 1)3 = 64 => 𝑎 + 𝑏 = => 𝑎 = − 𝑏 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => (3 − 𝑏)3 + 𝑏 = => 𝑏 = ℎ𝑎𝑦 𝑏 = 𝑣ớ𝑖 𝑏 = => 𝑎 = 𝑣ớ𝑖 𝑏 = => 𝑎 = 𝑡ừ đâ𝑦 𝑔𝑖ả𝑖 𝑟𝑎 đượ𝑐 𝑥, 𝑦 87) 𝑥 + = (2𝑥 + 1)√√𝑥 + + đặ𝑡 𝑡 = √√𝑥 + + 2(đ𝑘: 𝑡 ≥ 0) => 2(𝑡 − 2)2 = 2𝑥 + => (𝑡 − 2)2 = (2(𝑡 − 2)2 − 1)𝑡 => −2𝑡 + 𝑡 + 8𝑡 − 4𝑡 − 7𝑡 + = 𝑣ậ𝑦 𝑡 = 1(𝑛ℎậ𝑛)ℎ𝑎𝑦 𝑡 = −1−√5 (𝑙𝑜ạ𝑖)ℎ𝑎𝑦 𝑡 = −1+√5 (𝑛ℎậ𝑛) ℎ𝑎𝑦 𝑡 = 𝑣ớ𝑖 𝑡 = => √𝑥 + = −1(𝑝𝑡𝑣𝑛) 𝑣ớ𝑖 𝑡 = 𝑣ớ𝑖 𝑡 = −1+√5 1+√33 => √√𝑥 + + = −1+√5 => √√𝑥 + + = 1+√33 (𝑝𝑡𝑣𝑛) => 𝑥 = √33−15 32 88) 4√𝑥 + 𝑥 + = −𝑥 − 2𝑥 + 4𝑥 + 5𝑥 + => −𝑥 − 2𝑥 + 4𝑥 + 5𝑥 + − 4√𝑥 + 𝑥 + = đặ𝑡 𝑡 = √𝑥 + 𝑥 + => 𝜕𝑡 − 4𝑡 − 𝑥 − 2𝑥 + 4𝑥 + 5𝑥 + − 𝜕(𝑥 + 𝑥 + 1) = √𝑑𝑒𝑛𝑡𝑎 = √16 − 4𝜕[−𝑥 − 2𝑥 + 4𝑥 + 5𝑥 + − 𝜕(𝑥 + 𝑥 + 1)] 𝑣ậ𝑦 𝑡𝑎 𝑐ℎọ𝑛 𝜕 = 𝑙à 𝑠ố đẹ𝑝 𝑛ℎấ𝑡 => 𝑡 − 4𝑡 − 𝑥 − 2𝑥 + 3𝑥 + 4𝑥 = 𝑑𝑒𝑛𝑡𝑎 = 16 − 4(−𝑥 − 2𝑥 + 3𝑥 + 4𝑥) = 4(𝑥 − 1)2 (𝑥 + 2)2 ≥ 1+√33 (𝑛ℎậ𝑛) 𝑣ậ𝑦 √𝑥 + 𝑥 + = 4−2(𝑥−1)(𝑥+2) (𝑎) ℎ𝑎𝑦 √𝑥 + 𝑥 + = 4+2(𝑥−1)(𝑥+2) (𝑏) 𝑣ớ𝑖 𝑝𝑡(𝑎): √𝑥 + 𝑥 + = − 𝑥 − 𝑥 đặ𝑡 𝑘 = √𝑥 + 𝑥 + => 𝑘 − = 𝑥 + 𝑥 => 𝑘 + 𝑘 = => 𝑘 = −1−√21 (𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑘 = √21−1 (𝑛ℎậ𝑛) 𝑣ớ𝑖 𝑝𝑡(𝑏): √𝑥 + 𝑥 + = 𝑥 + 𝑥 đặ𝑡 𝑚 = √𝑥 + 𝑥 + 1(đ𝑘: 𝑚 ≥ 0) => 𝑚 = 𝑚2 − => 𝑚 = 𝑣ớ𝑖 𝑘 = 𝑣ớ𝑖 𝑚 = 1−√5 (𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑚 = √21−1 => √𝑥 + 𝑥 + = √21−1 1+√5 => √𝑥 + 𝑥 + = 1+√5 2 1+√5 => 𝑥 = => 𝑥 = −1±√19−2√21 −1±√3+2√5 𝑥 + 𝑦 + 𝑧 = 𝑝𝑖(1) 89) { 𝑡𝑎𝑛𝑥 𝑡𝑎𝑛𝑧 = 2(2) 𝑡𝑎𝑛𝑦 𝑡𝑎𝑛𝑧 = 18(3) đâ𝑦 𝑙à đề 𝑉𝑀𝑂 2009 𝑙ấ𝑦 𝑝𝑡(1)𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 𝑥 = 𝑝𝑖 − 𝑦 − 𝑧 => tan(𝑝𝑖 − 𝑦 − 𝑧) 𝑡𝑎𝑛𝑧 = => − tan(𝑦 + 𝑧) 𝑡𝑎𝑛𝑧 = − tan(𝑦 + 𝑧) 𝑡𝑎𝑛𝑧 = 2(∗) 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ: { 𝑡𝑎𝑛𝑦 𝑡𝑎𝑛𝑧 = 18(∗∗) 𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 𝑚à 𝑡𝑎 𝑐ó − tan(𝑦 + 𝑧) = 𝑡𝑎𝑛𝑦.𝑡𝑎𝑛𝑧−1 𝑡𝑎𝑛𝑦+𝑡𝑎𝑛𝑧 => {𝑡𝑎𝑛𝑦.𝑡𝑎𝑛𝑧−1 𝑡𝑎𝑛𝑧 = 2(∗) 𝑡𝑎𝑛𝑦 𝑡𝑎𝑛𝑧 = 18(∗∗) (𝑎+𝑏)𝑏 đặ𝑡 𝑎 = 𝑡𝑎𝑛𝑦, 𝑏 = 𝑡𝑎𝑛𝑧 => { 𝑎𝑏−1 𝑎𝑏 = 18(∗∗) 𝑣ớ𝑖 𝑡𝑎𝑛𝑧 = −4 => 𝑡𝑎𝑛𝑥 = − 𝑣ớ𝑖 𝑡𝑎𝑛𝑧 = => 𝑡𝑎𝑛𝑥 = = 2(∗) => { 𝑡𝑎𝑛𝑦 = − 𝑡𝑎𝑛𝑦 = ℎ𝑎𝑦 { 𝑡𝑎𝑛𝑧 = −4 𝑡𝑎𝑛𝑧 = 90) { 𝑦 + 𝑦 + 4(𝑥 − 𝑦 − 1) = 𝑥𝑦 (1) (𝑥 + 1)𝑦 + 𝑥 (2𝑦 + 1) = 𝑥 − 3𝑥 − 2(2) để ý 𝑝𝑡(1): 𝑦 (𝑦 + − 𝑥) + 4(𝑥 − 𝑦 − 1) = => (𝑦 − 4)(𝑦 + − 𝑥) = 𝑣ậ𝑦 𝑦 = ℎ𝑎𝑦 𝑦 + = 𝑥 𝑣ớ𝑖 𝑦 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) => 𝑝𝑡𝑣𝑛 𝑣ớ𝑖 𝑦 = −2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 𝑥 = −2 𝑣ớ𝑖 𝑦 = 𝑥 − 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): => 𝑥 + 𝑥 + = => 𝑥 − 2𝑥 + + 2𝑥 + 𝑥 + = => (𝑥 − 1)2 + (𝑥 + 4) + 91) { 15 = 0(𝑝𝑡𝑣𝑛) (𝑥 − 𝑥 + 2)𝑦 + 𝑥 = 0(1) (𝑥 − 4𝑥 − 1)𝑦 + (2𝑥 + 𝑥)𝑦 + 𝑥 = 0(2) để ý 𝑝𝑡(2): 𝑥 𝑦 − 4𝑥 𝑦 − 𝑦 + 2𝑥 𝑦 + 𝑥𝑦 + 𝑥 = => 𝑥 𝑦 + 2𝑥 𝑦 + 𝑥 − 4𝑥 𝑦 − 𝑦 + 𝑥𝑦 = => (𝑥 𝑦 + 𝑥)2 − 4𝑥 𝑦 − 𝑦 + 𝑥𝑦 = 𝑝𝑡(1): (𝑥 𝑦 + 𝑥) = 𝑥𝑦 − 2𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) => (𝑥𝑦 − 2𝑦)2 − 4𝑥 𝑦 − 𝑦 + 𝑥𝑦 = 𝑣ậ𝑦 𝑦 = ℎ𝑎𝑦 3𝑥 𝑦 + 4𝑥𝑦 − 𝑥 − 3𝑦 = 0(∗) 𝑣ớ𝑖 𝑦 = => 𝑥 = 𝑣ớ𝑖 3𝑥 𝑦 + 4𝑥𝑦 − 𝑥 − 3𝑦 = 0(∗) 𝑥 => 𝑦(3𝑥 + 4𝑥 − 3) = 𝑥 => 𝑦 = 3𝑥 +4𝑥−3 (𝑥 −𝑥+2)𝑥 𝑙ấ𝑦 𝑝𝑡(∗)𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 3𝑥 +4𝑥−3 + 𝑥 = => 𝑥 − 𝑥 + 2𝑥 + 3𝑥 + 4𝑥 − 3𝑥 = => 4𝑥 + 3𝑥 − 𝑥 = 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = −1 ℎ𝑎𝑦 𝑥 = 92) { 𝑥(𝑦 + 1) = 𝑦(𝑧 + 1) = 𝑧(𝑥 + 1)(1) 𝑥 + 2𝑦 + 3𝑧 + = 0(2) 𝑙ấ𝑦 𝑝𝑡(2) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): 𝑦 + = => 𝑥 ( −𝑥−3𝑧 => 𝑦 + = 𝑥2 ) = 𝑧(𝑥 + 1) => 𝑧 = − 5𝑥+2 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): −𝑥+ 3𝑥2 5𝑥+2 −𝑥 −6𝑥−2 => ( −𝑥−3𝑧 5𝑥+2 => 𝑦 = −𝑥 −6𝑥−2 5𝑥+2 𝑥2 𝑡ừ đâ𝑦 𝑡ℎế 𝑛𝑔ượ𝑐 𝑣à𝑜 𝑝𝑡(1): −𝑥 ) (− 5𝑥+2 + 1) = (5𝑥+2) (𝑥 + 1) 𝑣ậ𝑦 𝑥 = − ℎ𝑎𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = − − 1 √5 ℎ𝑎𝑦 𝑥 = − + √5 𝑣ớ𝑖 𝑥 = − => 𝑦 = − => 𝑧 = − 3 𝑣ớ𝑖 𝑥 = => 𝑦 = − => 𝑧 = − 3 √5 => 𝑦 = − √5 => 𝑦 = + 𝑣ớ𝑖 𝑥 = − − 𝑣ớ𝑖 𝑥 = − + √5 => 𝑧 = 1 √5 => 𝑧 = − − √5 −2 √5 93) √2𝑥 + + √17 − 2𝑥 = 𝑥 − 8𝑥 + 17𝑥 − 8𝑥 + 22 𝑡𝑎 𝑐ó 𝑉𝑃 = 𝑥 − 8𝑥 + 16𝑥 + 𝑥 − 8𝑥 + 16 + = 𝑥 (𝑥 − 4)2 + (𝑥 − 4)2 + = (𝑥 + 1)(𝑥 − 4)2 + => √2𝑥 + + √17 − 2𝑥 − = (𝑥 + 1)(𝑥 − 4)2 3√2𝑥+1 𝑡ℎ𝑒𝑜 𝑏đ𝑡 𝑐𝑎𝑢𝑐ℎ𝑦 𝑡ℎì: { 3√17−2𝑥 2𝑥+1+9 ≤ 3( )= 17−2𝑥+9 ≤ 3( 2𝑥+10 )= 26−2𝑥 => 𝑉𝑇 ≤ 𝑚à 𝑐ũ𝑛𝑔 𝑐ó 𝑉𝑇 ≥ 𝑣ậ𝑦 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 𝑥 = 3𝑥−𝑦 𝑥 + 𝑥 +𝑦 = 3(1) 94) { 𝑥+3𝑦 𝑦 − 𝑥 +𝑦 = 0(2) 𝑛ℎâ𝑛 𝑦 𝑣à𝑜 𝑝𝑡(1): 𝑥𝑦 + 𝑛ℎâ𝑛 𝑥 𝑣à𝑜 𝑝𝑡(2): 𝑥𝑦 − 3𝑥𝑦−𝑦 𝑥 +𝑦 𝑥 +3𝑦𝑥 𝑥 +𝑦 = 3𝑦(đ𝑘: 𝑦 ≠ 0) = 0(đ𝑘: 𝑥 ≠ 0) => 2𝑥𝑦 − = 3𝑦 𝑣ậ𝑦 𝑦(2𝑥 − 3) = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑥 + 2𝑥−3 𝑥2+ (2𝑥−3)2 3𝑥− 3 = (đ𝑘: 𝑥 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑛ê𝑛 𝑥 ≠ 2) 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = 0(𝑙𝑜ạ𝑖) ℎ𝑎𝑦 𝑥 = 𝑥+2𝑦 𝑥 + 𝑥 +𝑦 = 2(1) 94′) { 2𝑥−𝑦 𝑦 + 𝑥 +𝑦 = 0(2) 𝑡𝑎 𝑥é𝑡 𝑦 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑛ℎâ𝑛 𝑦 𝑣à𝑜 𝑝𝑡(1): 𝑥𝑦 + 𝑥𝑦+2𝑦 𝑥 +𝑦 = 2𝑦(đ𝑘: 𝑦 ≠ 0) 𝑡𝑎 𝑥é𝑡 𝑥 = 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑛ℎâ𝑛 𝑥 𝑣à𝑜 𝑝𝑡(2): 𝑥𝑦 + 2𝑥 −𝑥𝑦 𝑥 +𝑦 = 0(đ𝑘: 𝑥 ≠ 0) 𝑐ộ𝑛𝑔 𝑝𝑡 𝑣à𝑜: 2𝑥𝑦 + = 2𝑦 => 𝑥𝑦 + = 𝑦 => (𝑥 − 1)𝑦 = −1 => 𝑦 = 1−𝑥 (đ𝑘: 𝑥 ≠ 1) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1): => 𝑥 + 1−𝑥 𝑥2+ (1−𝑥)2 𝑥+ = => 𝑥 = ℎ𝑎𝑦 𝑥 = 𝑡ạ𝑖 𝑠𝑎𝑜 𝑡ô𝑖 𝑙ạ𝑖 𝑙ấ𝑦 𝑏à𝑖 𝑛à𝑦 để 𝑔𝑖ả𝑖 𝑏ở𝑖 𝑣ì 𝑡ừ 𝑡𝑟ướ𝑐 𝑘ℎ𝑖 𝑛ℎắ𝑐 đế𝑛 𝑏à𝑖 𝑛à𝑦 𝑡ℎì đề𝑢 𝑥à𝑖 𝑝ℎứ𝑐 ℎó𝑎 𝑣à 𝑡ơ𝑖 𝑙ạ𝑖 đặ𝑡 𝑣ấ𝑛 đề 𝑛ế𝑢 𝑘ℎơ𝑛𝑔 𝑥à𝑖 𝑝ℎứ𝑐 ℎó𝑎 𝑐ó đượ𝑐 𝑘ℎô𝑛𝑔 ? ? ? 𝑣à đâ𝑦 𝑙à 𝑘ế𝑡 𝑞𝑢ả 95) { 𝑥 − 3𝑥 + = √𝑦 + 3𝑦 (1) 3√𝑥 − = √𝑦 + 8𝑦(2) 𝑝𝑡(1): 𝑥 − 3𝑥 + = 𝑦√𝑦 + => 𝑥 − 3𝑥 + = (𝑦 + 3)√𝑦 + − 3√𝑦 + 3 => 𝑥 − 3𝑥 + 3𝑥 − − 3𝑥 + = (√𝑦 + 3) − 3(√𝑦 + 3) => (𝑥 − 1)3 − 3(𝑥 − 1) = (√𝑦 + 3) − 3(√𝑦 + 3) 𝑣ì ℎà𝑚 𝑛à𝑦 𝑘ℎơ𝑛𝑔 đồ𝑛𝑔 𝑏𝑖ế𝑛 𝑛ê𝑛 𝑡𝑎 𝑠ẽ 𝑡á𝑐ℎ 𝑡ℎử 𝑣ậ𝑦 𝑥 − = √𝑦 + ℎ𝑎𝑦 (𝑥 − 1)2 + (𝑥 − 1)√𝑦 + + 𝑦 = 𝑣ớ𝑖 (𝑥 − 1)2 + (𝑥 − 1)√𝑦 + + 𝑦 = 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑚ớ𝑖 𝑠𝑎𝑢: { (𝑥 − 1)2 + (𝑥 − 1)√𝑦 + + 𝑦 = 0(∗) 3√𝑥 − = √𝑦 + 8𝑦(∗∗) 𝑥≥2 𝑥≥2 Đ𝑘: {𝑦 ≤ −8 ℎ𝑎𝑦 𝑦 ≥ => { 𝑦≥0 𝑦 ≥ −3 𝑣ậ𝑦 (𝑥 − 1)2 + (𝑥 − 1)√𝑦 + + 𝑦 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0) 𝑣ớ𝑖 𝑥 − = √𝑦 + 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑠𝑎𝑢: { 𝑙ấ𝑦 𝑝𝑡(𝑎): 𝑥 − 2𝑥 − = 𝑦 (đ𝑘: { 𝑥 − = √𝑦 + 3(𝑎) 3√𝑥 − = √𝑦 + 8𝑦(𝑏) 𝑦 ≥ −3 ) 𝑥≥1 => 9𝑥 − 18 = 𝑦 + 8𝑦 = (𝑥 − 2𝑥 − 2)2 + 8(𝑥 − 2𝑥 − 2) => (𝑥 − 2𝑥 − 2)2 + 8𝑥 − 25𝑥 + = => 𝑥 + 4𝑥 + − 4𝑥 + 8𝑥 − 4𝑥 + 8𝑥 − 25𝑥 + = => 𝑥 − 4𝑥 + 8𝑥 − 17𝑥 + = => (𝑥 − 3)(𝑥 − 𝑥 + 5𝑥 − 2) = 𝑣ậ𝑦 𝑥 = 3(𝑡ℎõ𝑎) ℎ𝑎𝑦 𝑥 − 𝑥 + 5𝑥 − = 0(𝑚) 𝑣ớ𝑖 𝑝𝑡(𝑚): 𝑥 (𝑥 − 1) + 5(𝑥 − 1) + = 0(đ𝑘: 𝑥 ≥ 1) 𝑑ễ 𝑡ℎấ𝑦 𝑝𝑡(𝑚)𝑛à𝑦 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑉𝑇 ≥ > 𝑣ớ𝑖 𝑥 = => 𝑦 = 96) 2(1 − 𝑥)√𝑥 + 2𝑥 − = 𝑥 − 2𝑥 − => 𝑥 − 2𝑥 − − (2 − 2𝑥)√𝑥 + 2𝑥 − = đặ𝑡 𝑡 = √𝑥 + 2𝑥 − => 𝜕𝑡 − (2 − 2𝑥)𝑡 + 𝑥 − 2𝑥 − − 𝜕(𝑥 + 2𝑥 − 1) = 𝑐ℎọ𝑛 𝑠ố 𝜕: √𝑑𝑒𝑛𝑡𝑎 = √(2 − 2𝑥)2 − 4𝜕[𝑥 − 2𝑥 − − 𝜕(𝑥 + 2𝑥 − 1)] 𝑐ℎ𝑜 𝑥 = 100 𝑣ậ𝑦 𝜕 = 𝑝𝑡 𝑡𝑟ở 𝑡ℎà𝑛ℎ => 𝑡 − (2 − 2𝑥)𝑡 − 4𝑥 = 𝑑𝑒𝑛𝑡𝑎 = − 8𝑥 + 4𝑥 + 16𝑥 = 4𝑥 + 8𝑥 + = (2𝑥 + 2)2 ≥ 𝑣ậ𝑦 √𝑥 + 2𝑥 − = −2𝑥 ℎ𝑎𝑦 √𝑥 + 2𝑥 − = 𝑣ớ𝑖 √𝑥 + 2𝑥 − = −2𝑥(đ𝑘: 𝑥 ≤ 0) => −3𝑥 + 2𝑥 − = 0(𝑝𝑡𝑣𝑛) 𝑣ớ𝑖 √𝑥 + 2𝑥 − = => 𝑥 + 2𝑥 − = 𝑣ậ𝑦 𝑥 = −1 + √6 ℎ𝑎𝑦 𝑥 = −1 − √6 97) 2(𝑥 + 1)√𝑥 + 𝑥 = 𝑥 + 𝐶á𝑐ℎ 1: 𝑏ì𝑛ℎ 𝑝ℎươ𝑛𝑔 𝑙ê𝑛 ℎế𝑡 => 4(𝑥 + 1)2 (𝑥 + 𝑥) = (𝑥 + 7)2 => 4(𝑥 + 2𝑥 + 1) (𝑥 + 𝑥) = 𝑥 + 14𝑥 + 49 => (4𝑥 + 8𝑥 + 4) (𝑥 + 𝑥) = 𝑥 + 14𝑥 + 49 => (4𝑥 + 8𝑥 + 4)(𝑥 + 3) = 𝑥 + 14𝑥 + 49𝑥 => 4𝑥 + 12𝑥 + 8𝑥 + 24𝑥 + 4𝑥 + 12 = 𝑥 + 14𝑥 + 49𝑥 => 𝑥 − 4𝑥 + 6𝑥 − 16𝑥 + 25𝑥 − 12 = => (𝑥 − 1)2 (𝑥 − 3)(𝑥 + 𝑥 + 4) = 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑥 = 3 𝐶á𝑐ℎ 2: đặ𝑡 𝑡 = √𝑥 + 𝑥 => 𝑡 = 𝑥 + 𝑥 (đ𝑘: 𝑡 ≥ 0) => 2(𝑥 + 1)𝑡 = 𝑡 𝑥 + => 𝑡 𝑥 − (2𝑥 + 2)𝑡 + = 𝑑𝑒𝑛𝑡𝑎 = 4𝑥 + 8𝑥 + − 16𝑥 = (2𝑥 − 2)2 ≥ 3 𝑣ậ𝑦 √𝑥 + 𝑥 = 𝑥 ℎ𝑎𝑦 √𝑥 + 𝑥 = 2(đế𝑛 đâ𝑦 𝑐ó 𝑡ℎể 𝑔𝑖ả𝑖 đượ𝑐 𝑟ồ𝑖 ) 98) 3𝑥 + 𝑥 − 18 + (4𝑥 + 18)(√4 − 𝑥 − √𝑥 − 3) + 8√4 − 𝑥√𝑥 − = đặ𝑡 𝑡 = √4 − 𝑥 − √𝑥 − => 𝑡 = − 2√(4 − 𝑥)(𝑥 − 3)(đ𝑘: 𝑡 ≥ 0) => 3𝑥 + 𝑥 − 18 + (4𝑥 + 18)𝑡 + 4(1 − 𝑡 ) = => −4𝑡 + (4𝑥 + 18)𝑡 + 3𝑥 + 𝑥 − 14 = 𝑑𝑒𝑛𝑡𝑎 = (4𝑥 + 18)2 + 16(3𝑥 + 𝑥 − 14) = (8𝑥 + 10)2 ≥ 𝑣ậ𝑦 𝑡 = 𝑣ớ𝑖 𝑡 = 3𝑥+7 3𝑥+7 ℎ𝑎𝑦 𝑡 = 2−𝑥 => 2√4 − 𝑥 − 2√𝑥 − = 3𝑥 + 7(đ𝑘: ≤ 𝑥 ≤ 4) đặ𝑡 𝑓(𝑥) = 2√4 − 𝑥 − 2√𝑥 − 𝑓 ′ (𝑥) = −1 √4−𝑥 − √𝑥−3 𝑑ễ 𝑡ℎấ𝑦 𝑓 ′ (𝑥) > 𝑣ớ𝑖 𝑚ọ𝑖 𝑥 𝑡ℎ𝑢ộ𝑐 [3,4] 𝑣ậ𝑦 𝑚𝑎𝑥𝑓(𝑥) = −2 𝑡ạ𝑖 𝑥 = 𝑣à 𝑚𝑖𝑛𝑓(𝑥) = 𝑡𝑎𝑖 𝑥 = 𝑡ứ𝑐 − ≤ 𝑓(𝑥) ≤ 𝑚à 𝑐ó ∶ 16 ≤ 3𝑥 + ≤ 19 𝑣ậ𝑦 𝑡ℎì 𝑝𝑡 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑣ớ𝑖 2√4 − 𝑥 − 2√𝑥 − + 𝑥 − = => 2√4 − 𝑥 + (√𝑥 − − 1) = 𝑑ễ 𝑡ℎấ𝑦 𝑉𝑇 ≥ 𝑉𝑃 𝑣ậ𝑦 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 𝑥 = 𝑥 − 2𝑥 = 𝑦(1) 𝑦 + 2𝑦 = 𝑧(2) 99) { 𝑥 + 𝑦 + 𝑧 + + √𝑥 − = 0(3) 𝑑ễ 𝑡ℎấ𝑦 𝑝𝑡(1): (𝑥 − 1)2 = 𝑦 + 𝑝𝑡(2): (𝑦 + 1)2 = 𝑧 + 𝑝𝑡(3): (𝑥 − 1) + (𝑦 + 1) + (𝑧 + 1) + √𝑥 − = đặ𝑡 𝑎 = 𝑥 − 1, 𝑏 = 𝑦 + 1, 𝑐 = 𝑧 + 𝑎2 = 𝑏(∗) 𝑏 = 𝑐(∗∗) 𝑡ừ đâ𝑦 𝑐ó ℎệ 𝑚ớ𝑖: { 𝑎 + 𝑏 + 𝑐 + √𝑎 = 0(∗∗∗) 𝑙ấ𝑦 𝑝𝑡(∗) + 𝑝𝑡(∗∗): 𝑏 + 𝑐 = 𝑎2 + 𝑏 𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗∗∗): => 𝑎 + 𝑎2 + 𝑏 + √𝑎 = 𝑙ấ𝑦 𝑝𝑡(∗)𝑡ℎế 𝑣à𝑜 𝑝𝑡 𝑡𝑟ê𝑛: => 𝑎 + 𝑎2 + 𝑎4 + √𝑎 = 𝑣ậ𝑦 𝑎 = ℎ𝑎𝑦 𝑎3 √𝑎 + 𝑎√𝑎 + √𝑎 + = 0(𝑝𝑡𝑣𝑛) 𝑣ớ𝑖 𝑎 = => 𝑏 = => 𝑐 = 𝑣ậ𝑦 𝑥 = => 𝑦 = −1 => 𝑧 = −1 2𝑥 = 𝑦(𝑥 + 1)(1) 100) { 3𝑦 = 𝑧(𝑦 + 𝑦 + 1)(2) 4𝑧 = 𝑥(𝑧 + 𝑧 + 𝑧 + 1)(3) 𝑑ễ 𝑡ℎấ𝑦 𝑥 = 𝑦 = 𝑧 = 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑔𝑖ả 𝑠ử 𝑥 ≠ 0, 𝑦 ≠ 0, 𝑧 ≠ 𝑑ễ 𝑡ℎấ𝑦 𝑝𝑡(1): 𝑥 + > => 2𝑥 𝑦 > 𝑚à 𝑥 > 𝑛ê𝑛 𝑦 > 𝑐ũ𝑛𝑔 𝑡ươ𝑛𝑔 𝑡ự 𝑛ℎư 𝑝𝑡(3) 𝑡𝑎 𝑐ũ𝑛𝑔 𝑐ó đượ𝑐 𝑥 > 𝑡ừ đâ𝑦 𝑡𝑎 𝑡ℎ𝑢 đượ𝑐 đ𝑘 𝑙à ∶ 𝑥, 𝑦, 𝑧 > 𝑦(𝑥 + 1) ≥ 2𝑦√𝑥 = 2𝑦𝑥(∗) 𝑧(𝑦 + 𝑦 + 1) ≥ 3𝑧 √𝑦 = 3𝑧𝑦 (∗∗) 𝑡ℎ𝑒𝑜 𝑐𝑎𝑢𝑐ℎ𝑦 𝑡𝑎 𝑐ó ∶ { 𝑥(𝑧 + 𝑧 + 𝑧 + ≥ 4𝑥 √𝑧12 = 4𝑥𝑧 (∗∗∗) 𝑛ℎâ𝑛 𝑝𝑡(∗), 𝑝𝑡(∗∗), 𝑝𝑡(∗∗∗) 𝑛ê𝑛 𝑡𝑎 đượ𝑐 𝑉𝑃 ≥ 24𝑥 𝑦 𝑧 = 𝑉𝑇 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑐ℎỉ 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 ∶ 𝑥 = 𝑦 = 𝑧 = 101) { √𝑥 + √𝑦 + 2√2 + 𝑥 + 𝑦 + (√𝑥 − √𝑦) = 6(1) 4𝑥√𝑥 + + 9𝑦√𝑦 + = 5𝑥 + 15𝑦 + 15(2) để ý 𝑝𝑡(2)𝑙à 𝑑ạ𝑛𝑔 𝑆 𝑂 𝑆 𝑛ℎư𝑛𝑔 𝑣ấ𝑛 đề đâ𝑦 𝑙à 𝑡á𝑐ℎ 𝑙à𝑚 𝑠𝑎𝑜 để 𝑥𝑢ấ𝑡 ℎ𝑖ệ𝑛 𝑛ó ? ? => 5𝑥 + 15𝑦 + 15 − 4𝑥√𝑥 + − 9𝑦√𝑦 + = đặ𝑡 𝑡ℎử ℎệ 𝑠ố 𝑏ấ𝑡 đị𝑛ℎ ∶ 2 𝑡(𝑥 + 𝑛√𝑥 + 3) + 𝑘(𝑦 + 𝑙√𝑦 + 8) = => 𝑡([1 + 𝑛2 ]𝑥 + 3𝑛2 + 2𝑛𝑥√𝑥 + 3) + 𝑘([1 + 𝑙 ]𝑦 + 8𝑙 + 2𝑙𝑦√𝑦 + 8) = 𝑡(1 + 𝑛2 ) = 5(∗) 3𝑛2 𝑡 + 8𝑙 𝑘 = 15 2𝑛𝑡 = −4(∗∗) 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó ℎệ 𝑠𝑎𝑢: 2𝑙𝑘 = −9(𝑎) {𝑘(1 + 𝑙 ) = 15(𝑏) 2 𝑙ấ𝑦 𝑝𝑡(∗∗)𝑡ℎế 𝑣à𝑜 𝑝𝑡(∗): 𝑡 = − 𝑛 => − 𝑛 (1 + 𝑛2 ) = => −2 − 2𝑛2 = 5𝑛 => 𝑛 = −2 ℎ𝑎𝑦 𝑛 = − 𝑛 = −2 => 𝑡 = ℎ𝑎𝑦 𝑣ớ𝑖 𝑛 = − => 𝑡 = 𝑙ấ𝑦 𝑝𝑡(𝑎) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(𝑏): 𝑘 = − 2𝑙 => − 2𝑙 (1 + 𝑙 ) = 15 => −9 − 9𝑙 = 30𝑙 => 𝑙 = − ℎ𝑎𝑦 𝑙 = −3 𝑣ớ𝑖 𝑙 = − => 𝑘 = 27 ℎ𝑎𝑦 𝑣ớ𝑖 𝑙 = −3 => 𝑘 = 𝑛 = −2 𝑡=4 𝑡ừ đâ𝑦 𝑡𝑎 𝑐ó 𝑐á𝑐 ℎệ 𝑠ố 𝑝ℎù ℎợ𝑝 ℎệ 𝑠𝑎𝑢 𝑙à ∶ 𝑙 = − {𝑘= 27 𝑡ừ đâ𝑦 𝑙ắ𝑝 𝑣à𝑜 𝑝𝑡 𝑡𝑟ê𝑛: => (𝑥 − √𝑥 +3 2 ) + 27 (𝑦 − √𝑦 +8 ) =0 𝑣ậ𝑦 𝑡𝑎 𝑐ó đượ𝑐 𝑉𝑇 ≥ 𝑉𝑃 𝑛ê𝑛 𝑑ấ𝑢 𝑏ằ𝑛𝑔 𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖: { 2𝑥 = √𝑥 + 3𝑦 = √𝑦 + 𝑣ậ𝑦 𝑥 = 1, 𝑦 = 1(𝑡ℎế 𝑣à𝑜 𝑝𝑡(1)𝑡ℎõ𝑎) ℎ𝑎𝑦 𝑥 = −1, 𝑦 = −1 (𝑙𝑜ạ𝑖 𝑑𝑜 𝑥, 𝑦 ≥ 0) 102) 𝑥 = √2 + √2 − √2 + 𝑥 đâ𝑦 𝑙à 𝑠𝑙𝑡 𝑐ủ𝑎 𝑡ℎầ𝑦 𝑂𝑟𝑙𝑎𝑛𝑑𝑜 𝐼𝑟𝑎ℎ𝑜𝑙𝑎 𝑂𝑟𝑡𝑒𝑔𝑎 𝑏ằ𝑛𝑔 𝑐á𝑐ℎ 𝑙ượ𝑛𝑔 𝑔𝑖á𝑐 𝑐ự𝑐 đẹ𝑝 ‼! đặ𝑡 𝑥 = 2𝑐𝑜𝑠𝑡(đ𝑘: − ≤ 𝑥 ≤ 1) => 𝑐𝑜𝑠𝑡 = √2 + √2 − √2 + 2𝑐𝑜𝑠𝑡 => 2𝑐𝑜𝑠𝑡 = √2 + √2 − √2(1 + 𝑐𝑜𝑠𝑡) 𝑡 => 2𝑐𝑜𝑠𝑡 = √2 + √2 − √2.2 cos (2) 𝑡 => 2𝑐𝑜𝑠𝑡 = √2 + √2 − cos (2) 𝑡 => 2𝑐𝑜𝑠𝑡 = √2 + √2(1 − cos (2) 𝑡 𝑡 => 2𝑐𝑜𝑠𝑡 = √2 + √2.2 sin2 (4) = √2 + sin (4) 𝑡 𝑡 => 4cos2 𝑡 = + sin (4) => 𝑐𝑜𝑠2𝑡 = sin (4) 𝑝𝑖 𝑡 => sin ( + 2𝑡) = sin (4) => 𝑥 = cos (9 𝑝𝑖)