𝑐á𝑐 𝑏à𝑖 ℎệ phương trình ℎ𝑎𝑦 𝑣à 𝑘ℎó (Nguyễn Trường Phát TP.HCM) 𝑥 + 𝑥 √𝑥 − 3√𝑥 − 2𝑥𝑦 + 6𝑦 − = 0(1) 𝑏à𝑖 1: { √𝑥 + 2𝑦 + + √𝑥 − 2𝑦 + = 3(2) 𝑔𝑖ả𝑖 𝑥 + 2𝑦 + ≥ đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑥≥0 𝑝𝑡(1): (√𝑥) + (√𝑥) − 3√𝑥 − − 2𝑥𝑦 + 6𝑦 = 𝑥(√𝑥 + 1) − 3(√𝑥 + 1) − 2𝑦(𝑥 − 3) = (𝑥 − 3)(√𝑥 + 1) − 2𝑦(𝑥 − 3) = 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 √𝑥 + − 2𝑦 = (∗∗)𝑣ớ𝑖 𝑥 = 𝑡ℎế 𝑣à𝑜 (2)𝑡ℎì 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦 = 16 𝑣à 𝑦 = − 𝑡ℎế 𝑥 = 𝑣à 𝑦 = 16 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎 𝑡ℎế 𝑥 = 𝑣à 𝑦 = − 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑐ũ𝑛𝑔 𝑡ℎõ𝑎 (∗)𝑣ớ𝑖 √𝑥 − 2𝑦 + = 𝑡ℎế 𝑣à𝑜 (2) √𝑥 + √𝑥 + + √𝑥 − √𝑥 + = 3(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1) 𝑡ℎế 𝑣à𝑜 (∗)𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦 = 𝑣ớ𝑖 𝑥 = 1, 𝑦 = 𝑡ℎế 𝑣à𝑜 (1) 𝑐ũ𝑛𝑔 𝑡ℎõ𝑎 𝑣ậ𝑦 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑛ℎư 𝑠𝑎𝑢: (1,1); (3,16); (3, − ) 2𝑥 + = 𝑦 + 4𝑦√𝑥(1) 𝑏à𝑖 2: { 𝑦 + √𝑥 + = 𝑦√𝑥(2) 𝑔𝑖ả𝑖 đ𝑘 để ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑥 ≥ để ý 𝑝𝑡(2): 𝑦 + = 𝑦√𝑥 − √𝑥 𝑙ấ𝑦(2)𝑡ℎế 𝑣à𝑜 (1) 𝑡𝑎 𝑐ó đượ𝑐: 2𝑥 − 4𝑦√𝑥 = (𝑦 + 1)(𝑦 − 1) = (𝑦√𝑥 − √𝑥)(𝑦 − 1) √𝑥 = ℎ𝑎𝑦 2√𝑥 − 4𝑦 = 𝑦 − 2𝑦 + √𝑥 = ℎ𝑎𝑦 2√𝑥 = 𝑦 + 2𝑦 + 𝑣ớ𝑖 𝑥 = 𝑡ℎì 𝑦 = ±1 𝑡ℎế 𝑣à𝑜 (2)𝑡𝑎 𝑛ℎậ𝑛 𝑥 = 0, 𝑦 = −1 𝑦 + 2𝑦 + 𝑣ớ𝑖 √𝑥 = 𝑡ℎế 𝑣à𝑜 (2) 𝑦 + 2𝑦 + 𝑦 + 2𝑦 + 𝑦+ +1 = 𝑦( ) 2 𝑦 + 4𝑦 + = 𝑦 + 2𝑦 + 𝑦 𝑦 = −1 𝑡ℎì 𝑥 = ℎ𝑎𝑦 𝑦 = ±√3, 𝑥 = ± 4√3 𝑣ớ𝑖 𝑦 = −1 𝑣à 𝑥 = 𝑡ℎế 𝑛𝑔ượ𝑐 𝑙ạ𝑖 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎 𝑣ớ𝑖 𝑦 = √3 𝑣à 𝑥 = + 4√3 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎 𝑣ớ𝑖 𝑦 = −√3 𝑣à 𝑥 = − 4√3 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑘ℎô𝑛𝑔 𝑡ℎõ𝑎 𝑣ậ𝑦 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑙ầ𝑛 𝑙ượ𝑡 𝑙à: (0, −1); (√3, + 4√3) √3𝑥 − + 4(2𝑥 + 1) = √𝑦 − + 3𝑦(1) 𝑏à𝑖 3: { (𝑥 + 𝑦)(2𝑥 − 𝑦) + 6𝑥 + 3𝑦 + = 0(2) 𝑔𝑖ả𝑖 đ𝑘 để ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑦≥1 𝑥≥ 𝑝𝑡(2): 2𝑥 − 𝑥𝑦 + 2𝑥𝑦 − 𝑦 + 6𝑥 + 3𝑦 + = 2𝑥 + 𝑥𝑦 − 𝑦 + 6𝑥 + 3𝑦 + = 2𝑥 + 𝑥(𝑦 + 6) − 𝑦 + 3𝑦 + = 𝑑𝑒𝑛𝑡𝑎 = 𝑦 + 12𝑦 + 36 − 8(−𝑦 + 3𝑦 + 4) = 9𝑦 − 12𝑦 + = (3𝑦 − 2)2 ≥0 −𝑦 − + 3𝑦 − 2𝑦 − 𝑦 − 𝑥= = = 4 𝑣ậ𝑦 [ −𝑦 − − 3𝑦 + −4𝑦 − 𝑥= = = −𝑦 − 4 𝑣ớ𝑖 𝑥 = 𝑦−4 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑦 = 12, 𝑥 = 𝑣ớ𝑖 𝑥 + 𝑦 = −1 𝑡ℎì 𝑣ơ 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑥 + 𝑦 > 𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (4,12) 𝑏à𝑖 4: { (𝑥 + 2)√𝑦 + − (𝑦 + 2)√𝑥 + = 0(1) 𝑥 − 𝑦 − 3𝑥 𝑦 − 5𝑦 + = 0(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 𝑦 ≥ −1 𝑝𝑡(1): (𝑥 + 4𝑥 + 4)(𝑦 + 1) = (𝑦 + 4𝑦 + 4)(𝑥 + 1) 𝑥 𝑦 + 𝑥 + 4𝑥 𝑦 + 4𝑥 + 4𝑦 + = 𝑦 𝑥 + 𝑦 + 4𝑦𝑥 + 4𝑦 + 4𝑥 + 𝑥 4𝑦 + 𝑥 = 𝑥 2𝑦2 + 𝑦2 𝑥 𝑦(𝑥 − 𝑦) = (𝑦 − 𝑥 )(𝑦 + 𝑥 ) 𝑣ậ𝑦 𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 𝑦 = −(𝑦 + 𝑥 ) 𝑣ớ𝑖 𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 (2): −3𝑦 − 5𝑦 + = 0[ 𝑦=1 𝑦=− 𝑥2 𝑣ớ𝑖 𝑥 𝑦 + 𝑦 = −𝑥 => 𝑦 = − 𝑡ℎế 𝑣à𝑜 (2) 𝑥 +1 2 𝑥2 𝑥2 𝑥2 𝑥 −( ) + 3𝑥 ( ) + 5( )+8= 𝑥 +1 𝑥 +1 𝑥 +1 𝑥 (𝑥 + 1)2 − 𝑥 + 3𝑥 (𝑥 + 1) + 5𝑥 (𝑥 + 1) + 8(𝑥 + 1)2 = 𝑥 (𝑥 + 2𝑥 + 1) − 𝑥 + 3𝑥 + 3𝑥 + 5𝑥 + 5𝑥 + 8(𝑥 + 1)2 = 𝑥 + 2𝑥 + 3𝑥 + 8𝑥 + 5𝑥 + 8(𝑥 + 2𝑥 + 1) = 𝑥 + 5𝑥 + 16𝑥 + 21𝑥 + = (𝑝𝑡 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑥 + 5𝑥 + 16𝑥 + 21𝑥 + > 0) 2𝑥 + 𝑥 + √𝑥 + = 2𝑦 + 𝑦 + √2𝑦 + 1(1) 𝑏à𝑖 5: { 𝑥 + 2𝑦 − 2𝑥 + 𝑦 − = 0(2) 𝑔𝑖ả𝑖 𝑥 ≥ −2 đ𝑘 để 𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑦≥− 𝑝𝑡(1) − 𝑝𝑡(2): 𝑥 − 2𝑦 + 3𝑥 − 𝑦 + + √𝑥 + = 2𝑦 + 𝑦 + √2𝑦 + (𝑥 + 1)2 + (𝑥 + 1) + √(𝑥 + 1) + = (2𝑦)2 + 2𝑦 + √2𝑦 + 𝑥é𝑡 𝑓(𝑡) = 𝑡 + 𝑡 + √𝑡 + 𝑣ậ𝑦 𝑓 ′ (𝑡) = 2𝑡 + + √𝑡 + 𝑡ứ𝑐 𝑓(𝑥 + 1) = 𝑓(2𝑦)𝑣ậ𝑦 𝑥 + = 2𝑦(3) >0 𝑙ấ𝑦 (3)𝑡ℎế (2): (2𝑦 − 1)2 + 2𝑦 − 2(2𝑦 − 1) + 𝑦 − = 4𝑦 − 4𝑦 + + 2𝑦 − 4𝑦 + + 𝑦 − = 6𝑦 − 7𝑦 + = [ 𝑦=1 𝑦= 𝑣ớ𝑖 𝑦 = 𝑣ậ𝑦 𝑥 = 𝑣ớ𝑖 𝑦 = 𝑣ậ𝑦 𝑥 = − 𝑥 𝑦√𝑥 + 𝑥 𝑦 = 2𝑥 √𝑥 + 2𝑥 𝑦(1) 𝑏à𝑖 6: { 𝑦√𝑥 (√2𝑥 − − 1) = √5(2𝑥 − 6)(2) 𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥 𝑦(√𝑥 + 𝑦) = 2𝑥 (√𝑥 + 𝑦) 𝑣ậ𝑦 √𝑥 = −𝑦 ℎ𝑎𝑦 𝑥 = ℎ𝑎𝑦 𝑦 = 2𝑥 𝑥 − 3𝑦 − 3𝑥 𝑦 + 𝑥𝑦 + 𝑥 = 3𝑦(1) 𝑏à𝑖 7: 3 3𝑥 + 36𝑦 − = 𝑥 √27𝑦 + { 2𝑥 + (2) 𝑥 𝑔𝑖ả𝑖 𝑝𝑡(1): (𝑥 − 3𝑦)(𝑥 + 𝑦 ) = −(𝑥 − 3𝑦) 𝑣ậ𝑦 𝑥 = 3𝑦 ℎ𝑎𝑦 𝑥 + 𝑦 = −1(𝑝𝑡 𝑣𝑛 𝑑𝑜 𝑥 + 𝑦 ≥ 0) 𝑣ớ𝑖 𝑥 = 3𝑦 => 𝑦 = 𝑥 𝑡ℎế (2): 3𝑥 + 4𝑥 − = 𝑥 √𝑥 + + (𝑙𝑖ê𝑛 ℎợ𝑝 ) 𝑥 3 2𝑥(𝑥 + 3) − 𝑦(𝑦 + 3) = 3𝑥𝑦(𝑥 − 𝑦)(1) 𝑏à𝑖 8: { (𝑥 − 2)2 = 4(2 − 𝑦)(2) 𝑔𝑖ả𝑖 𝑝𝑡(1)2𝑥 + 6𝑥 − 𝑦 − 3𝑦 = 3𝑥 𝑦 − 3𝑥𝑦 2𝑥 − 3𝑥 𝑦 + 3𝑥𝑦 − 𝑦 + 6𝑥 − 3𝑦 = (2𝑥 − 𝑦)(𝑥 − 𝑥𝑦 + 𝑦 ) + 3(2𝑥 − 𝑦) = 2𝑥 = 𝑦 ℎ𝑎𝑦 𝑥 − 𝑥𝑦 + 𝑦 = −3 (𝑝𝑡 𝑣𝑛) 𝑣ờ𝑖 2𝑥 = 𝑦 𝑡ℎế 𝑣à𝑜 (2): (𝑥 − 2)2 = 4(2 − 2𝑥) 𝑥 − 4𝑥 + = − 8𝑥 (𝑥 + 2𝑥 − 2)(… ) = => 𝑥 = −1 ± √3 𝑏à𝑖 9: { √𝑥 + + √−2 − 𝑥 + √−1 − 10𝑦 = 8(1) 𝑥√𝑦 − 4𝑦 + + √𝑥 + = 𝑦√𝑥 + − √𝑦 − 4𝑦 + 5(2) (đã 𝑔𝑖ả𝑖 𝑟ồ𝑖 ‼!) 2(2𝑥 − 1)3 + 2𝑥 = (2𝑦 − 1)√𝑦 − + 1(1) 𝑏à𝑖 10: { √4𝑥 − + √2𝑦 − = 2√2(2) 𝑔𝑖ả𝑖 𝑦≥1 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑥≥ 𝑝𝑡(1): 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ℎ𝑜 1 8𝑥 − 12𝑥 + 6𝑥 − + 𝑥 = (𝑦 − ) √𝑦 − + 2 8𝑥 − 12𝑥 + 6𝑥 − (2𝑥 − 1)3 + 3 √𝑦 − = (√𝑦 − 1) + 2 2𝑥 − √𝑦 − = (√𝑦 − 1) + 2 𝑥é𝑡 𝑓(𝑡) = 𝑡 + 𝑡 => 𝑓 ′ (𝑡) = 3𝑡 + > 2 𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 𝑓(2𝑥 − 1) = 𝑓(√𝑦 − 1) 𝑣ậ𝑦 2𝑥 − = √𝑦 − 1(𝑡ℎế 𝑣à𝑜 (2)) 4𝑥 − 4𝑥 + = 𝑦 − 𝑦 = 4𝑥 − 4𝑥 + √4𝑥 − + √2(4𝑥 − 4𝑥 + 2) − = 2√2 √4𝑥 − + √8𝑥 − 8𝑥 + = 2√2(𝑙𝑖ê𝑛 ℎợ𝑝 𝑥 = 𝑙ư𝑢 ý 𝑠𝑎𝑢 𝑘ℎ𝑖 𝑙𝑖ê𝑛 ℎợ𝑝 𝑛ℎớ 𝑏ấ𝑚 𝑛𝑔ℎ𝑖ệ𝑚 𝑙ạ𝑖 𝑙ầ𝑛 𝑛ữ𝑎 𝑡𝑟á𝑛ℎ 𝑛𝑔ℎ𝑖ệ𝑚 𝑘é𝑝) 𝑥 = 𝑡ℎì 𝑦 = 𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (1,2) (𝑥 − 𝑦)(𝑥 + 𝑦 + 𝑦 ) = 𝑥(𝑦 + 1)(1) 𝑏à𝑖 11: { (𝑦 + 2)2 √𝑥 + 4𝑥 = + (2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑥 ≥ 𝑝𝑡(1): 𝑥 + 𝑥𝑦 + 𝑥𝑦 − 𝑥𝑦 − 𝑦 − 𝑦 = 𝑥𝑦 + 𝑥 𝑥 + 𝑥𝑦 − 𝑦 − 𝑦 = 𝑥𝑦 + 𝑥 𝑥(𝑥 + 𝑦 ) = 𝑥(𝑦 + 1) + 𝑦 (𝑦 + 1) 𝑥(𝑥 + 𝑦 ) = (𝑥 + 𝑦 )(𝑦 + 1) 𝑣ậ𝑦 𝑥 + 𝑦 = ℎ𝑎𝑦 𝑥 = 𝑦 + 𝑣ớ𝑖 𝑥 + 𝑦 = (𝑝𝑡𝑣𝑛)𝑑𝑜 𝑥 + 𝑦 > 𝑣ớ𝑖 𝑦 = 𝑥 − 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) √𝑥 (𝑥 + 1)2 𝑥 + 2𝑥 + + 4𝑥 = + =1+ 3 3√𝑥 + 4𝑥 = 𝑥 + 2𝑥 + 4(𝑙𝑖ê𝑛 ℎợ𝑝 𝑣ớ𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 2) 𝑣ớ𝑖 𝑥 = 𝑡ℎì 𝑦 = 𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: (2,1) 𝑥 + 𝑦 − 6𝑥 + 15𝑥 + 3𝑦 − 14 = 0(1) 𝑏à𝑖 12: { 4 √𝑥 + √𝑦 + √𝑥 + √𝑦 = 4(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑥, 𝑦 ≥ á𝑝 𝑑ụ𝑛𝑔 𝐵Đ𝑇 𝑏𝑢ℎ𝑖𝑎, 𝑑ấ𝑢=𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 𝑥 = 𝑦 = 𝑡ℎế 𝑣à𝑜 (1)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎 𝑣ậ𝑦 ℎ𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ (1,1) + (𝑥𝑦)3 = 19𝑥 (1) 𝑏à𝑖 13: { 𝑦 + 𝑥𝑦 = −6𝑥 (2) 𝑔𝑖ả𝑖 𝑇𝐻1: 𝑥 = 𝑡ℎế 𝑣à𝑜 (1) 𝑝𝑡𝑣𝑛 𝑇𝐻2: 𝑥 ≠ 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 𝑥 : + 𝑦 = 19 𝑥 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(2)𝑐ℎ𝑜 𝑥 : 𝑦 𝑦2 + = −6 𝑥2 𝑥 𝑡𝑎 𝑐ó − 𝑝𝑡(1) − 19 𝑝𝑡(2) = 19𝑦 19𝑦 − − 6𝑦 − − =0 𝑥 𝑥 𝑥 2𝑥 [ 𝑦=− 𝑥 𝑦=− 3𝑥 𝑦=− 16(𝑥𝑦)3 − 9𝑦 = (2𝑥𝑦 − 𝑦)(4𝑥𝑦 + 3)(1) 𝑏à𝑖 14: { 4𝑥 𝑦 + 2𝑥𝑦 + 𝑦 = 3(2) 𝑔𝑖ả𝑖 𝑡ℎế (2) 𝑣à𝑜 (1): 16𝑥 𝑦 − 9𝑦 = 𝑦 (2𝑥 − 1)(4𝑥 + 6𝑥 + 1) 𝑣ậ𝑦 𝑦 = ℎ𝑎𝑦 16𝑥 − = 8𝑥 + 12𝑥 + 2𝑥 − 4𝑥 − 6𝑥 − 𝑦 = ℎ𝑎𝑦 8𝑥 − 8𝑥 + 4𝑥 − = + 6𝑦 = 𝑏à𝑖 15: 𝑥 − √𝑥 − 2𝑦(1) 𝑦 √𝑥 + √𝑥 − 2𝑦 = 𝑥 + 3𝑦 − 2(2) { 𝑔𝑖ả𝑖 𝑦≠0 𝑥 − 2𝑦 ≥ đ𝑘 để ℎ𝑝𝑡 𝑡𝑟ê𝑛 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ { 𝑥 + √𝑥 − 2𝑦 ≥ 𝑝𝑡(1): 2𝑦 + 6𝑦 = 𝑥 − 𝑦√𝑥 − 2𝑦 đặ𝑡 𝑡ℎử ℎệ 𝑠ố ∶ 𝑥 − 2𝑦 + 𝑎𝑏 + (𝑎 + 𝑏)√𝑥 − 2𝑦 = { 𝑥 − 2𝑦 + 𝑎𝑏 = 𝑥 − 2𝑦 − 6𝑦 𝑎 + 𝑏 = −𝑦 (−𝑦)𝑏 − 𝑏 + 6𝑦 = 0(∗) { 𝑎 = (−𝑦) − 𝑏 𝑑𝑒𝑛𝑡𝑎(∗) = 𝑦 + 24𝑦 = (5𝑦)2 ≥ 𝑦 − 5𝑦 √𝑥 − 2𝑦 = −2 = 2𝑦 [ 𝑦 + 5𝑦 √𝑥 − 2𝑦 = −2 = −3𝑦 (𝑥 + 𝑦)(𝑥 + 4𝑦 + 𝑦) + 3𝑦 = 0(1) 𝑏à𝑖 16: { √𝑥 + 2𝑦 + − 𝑦 + 𝑦 + = 0(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ 𝑥 + 2𝑦 ≥ −1 để ý 𝑝𝑡(1): 𝑥 + 4𝑥𝑦 + 2𝑥𝑦 + 4𝑦 + 𝑦 + 3𝑦 = 𝑥 + 𝑥(4𝑦 + 2𝑦) + 4𝑦 + 𝑦 + 3𝑦 = 𝑑𝑒𝑛𝑡𝑎 = 16𝑦 + 16𝑦 + 4𝑦 − 16𝑦 − 4𝑦 − 12𝑦 = 4𝑦 ≥ −4𝑦 − 2𝑦 − 2𝑦 = −3𝑦 − 𝑦 [ −4𝑦 − 2𝑦 + 2𝑦 𝑥= = −𝑦 − 𝑦 𝑥= (∗)𝑣ớ𝑖 𝑥 = −3𝑦 − 𝑦 𝑡ℎế 𝑣à𝑜 (2): √−𝑦 − 𝑦 + − 𝑦 + 𝑦 + = 0( 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑦 = −1) 𝑣ậ𝑦 𝑦 = −1 𝑡ℎì 𝑥 = −2 𝑡ℎế 𝑣à𝑜 đ𝑘 𝑡ℎấ𝑦 𝑡ℎõ𝑎 (∗∗)𝑣ớ𝑖 𝑥 = −𝑦 − 𝑦 𝑡ℎế 𝑣à𝑜 (2): √𝑦 − 𝑦 + − 𝑦 + 𝑦 + = 𝑥 𝑥 𝑥 𝑥 𝑥 𝑝𝑡(1): √( ) + ( ) + = − ( ) 𝑦 𝑦 𝑦 𝑦 𝑦 đặ𝑡 𝑎 = 𝑥 đư𝑎 𝑣ề 𝑝𝑡 𝑣ô 𝑡ỷ 𝑡ℎ𝑒𝑜 𝑎: 𝑦 √𝑎3 + 𝑎2 + 𝑎 = 𝑎 − 𝑎2 (𝑣ậ𝑦 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑎 = 𝑣à 𝑎 = −1) 𝑎 = 𝑡ℎì 𝑥 = 𝑡ℎế 𝑡ℎử 𝑣à𝑜 𝑝𝑡(2): (𝑝𝑡𝑣𝑛) 𝑎 = −1 𝑡ℎì 𝑥 = −𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): √5𝑥 − + √9 − 𝑥 = 2𝑥 + 3𝑥 − 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑥 = 8𝑦 − + √𝑥 = √2𝑦(−3𝑥 + 5𝑥 − 2)(1) 𝑏à𝑖 64: { √𝑥 + 5𝑦 − √9𝑦 − 𝑥 = 4𝑦 − 𝑥(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 0, 𝑥 ≥ (𝑐ũ𝑛𝑔 𝑡ươ𝑛𝑔 𝑡ự 𝑛ℎư 𝑏à𝑖 𝑡𝑟ê𝑛)𝑇𝐻1: 𝑛ế𝑢 𝑦 = 𝑡ℎì 𝑥 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) 𝑡ℎấ𝑦 𝑘ℎô𝑛𝑔 𝑡ℎõ𝑎 𝑇𝐻2: 𝑛ế𝑢 𝑦 ≠ 𝑡ℎì 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(2)𝑐ℎ𝑜 𝑦: 𝑥 𝑥 𝑥 √( ) + − √9 − ( ) = − 𝑦 𝑦 𝑦 đặ𝑡 𝑎 = 𝑥 𝑣ậ𝑦 𝑔𝑖ả𝑖 𝑝𝑡 𝑣ô 𝑡ỷ 𝑡ℎ𝑒𝑜 𝑎: 𝑦 √𝑎2 + − √9 − 𝑎3 = − 𝑎 (𝑝𝑡 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑎 = 2) 𝑣ậ𝑦 𝑥 = 2𝑦 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) (𝑑à𝑛ℎ 𝑐ℎ𝑜 𝑏ạ𝑛 đọ𝑐 ‼) 𝑥 + 𝑥 = 𝑦 − 𝑦(1) 𝑏à𝑖 65: { 𝑦 + 𝑦 = 𝑥 − 𝑥(2) 𝑔𝑖ả𝑖 { 𝑥(𝑥 + 1) = 𝑦(𝑦 − 1)(𝑦 + 1)(1) 𝑦(𝑦 + 1) = 𝑥(𝑥 − 1)(𝑥 + 1)(2) 𝑙ấ𝑦 𝑝𝑡(1) 𝑝𝑡(2): 𝑥𝑦(𝑥 + 1)(𝑦 + 1) = 𝑥𝑦(𝑥 + 1)(𝑦 + 1)(𝑥 − 1)(𝑦 − 1) 𝑥=0 𝑦=0 𝑣ậ𝑦 [ 𝑥 = −1 𝑦 = −1 (𝑥 − 1)(𝑦 − 1) = 𝑣ớ𝑖 𝑥 = 𝑡ℎì 𝑦 = 𝑣à 𝑦 = ±1 𝑣ớ𝑖 𝑦 = 𝑡ℎì 𝑥 = 𝑣à 𝑥 = −1 𝑣à 𝑥𝑦 − 𝑥 − 𝑦 = 𝑙ấ𝑦 𝑝𝑡(1) + 𝑝𝑡(2): 𝑥 + 𝑦 + 2(𝑥 + 𝑦) = (𝑥 + 𝑦)(𝑥 − 𝑥𝑦 + 𝑦 ) 𝑥𝑦 − (𝑥 + 𝑦) = { (𝑥 + 𝑦)2 − 2(𝑥 + 𝑦) − 2𝑥𝑦 = (𝑥 + 𝑦)((𝑥 + 𝑦)2 − 3𝑥𝑦) 3√4(𝑥 + 𝑥) + 2√2𝑦 − 𝑦 = 3(𝑥 + 𝑦) + 2(1) 𝑏à𝑖 66: { 2016√𝑦 − + 𝑦 + = 𝑥 + 2𝑦(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ (𝑑𝑜 𝑝𝑡(2)) 𝑣à 𝑦 ≥ 𝑝𝑡(1): 3√4𝑥(𝑥 + 1) + 2√2𝑦 − 𝑦 ≤ + 2𝑥 + 𝑥 + + 𝑦 + 2𝑦 − = 3(𝑥 + 𝑦) + 𝑥=1 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑡𝑎 𝑡ℎấ𝑦 𝑡ℎõ𝑎) 𝑑ấ𝑢 "=" xảy : { 𝑦=1 4√𝑥 + + 𝑥𝑦√4 + 𝑦 = 0(1) 𝑏à𝑖 67: { 𝑦4 − 𝑥 +𝑥 =𝑦 +𝑦 + (2) 𝑔𝑖ả𝑖 𝑥𝑦 ≥ đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à ∶ { 𝑥 ≥ −1 để ý 𝑝𝑡(1): 16𝑥 + 16 = 4(𝑥𝑦)2 + 𝑥 𝑦 (4 − 𝑥𝑦 )(4 + 𝑥𝑦 ) + 4𝑥(4 − 𝑥𝑦 ) = 𝑣ậ𝑦 𝑥𝑦 = ℎ𝑎𝑦 + 𝑥𝑦 + 4𝑥 = 𝑥𝑦 = ℎ𝑎𝑦 𝑥(4 + 𝑦 ) = −4 => 𝑦 = − − 4(𝑣𝑠 𝑥 ≠ 0) 𝑥 4𝑥𝑦 + (𝑥𝑦 − 2)2𝑥𝑦 + 𝑥𝑦 = 3(1) 𝑏à𝑖 68: { (𝑥 + 𝑥𝑦 + 2)(3 𝑥 + 8𝑥 + 12𝑥 + 𝑥𝑦 + 20) = 5(𝑥 + 𝑥𝑦)3 (2) 𝑔𝑖ả𝑖 𝑝𝑡(1): (2𝑥𝑦 )2 + (𝑥𝑦 − 2) 2𝑥𝑦 + (𝑥𝑦 − 2) − = (2𝑥𝑦 − 1)(2𝑥𝑦 + 1) + (𝑥𝑦 − 2)(2𝑥𝑦 + 1) = 𝑣ậ𝑦 2𝑥𝑦 + 𝑥𝑦 − = ℎ𝑎𝑦 2𝑥𝑦 = −1(𝑙𝑜ạ𝑖) 𝑥𝑦 = (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) 𝑥√4𝑦 + + 2𝑦√𝑥 + = 0(1) 𝑏à𝑖 69: {2(1 + 23𝑥−4𝑦 ) 2.23𝑥 + = 2−2𝑦 (1 + 2−2𝑦 + 22𝑥 )(2) + 2−2𝑦 + 2−4𝑦 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ 0, 𝑦 ≥ 4(𝑥𝑦)2 + 𝑥 = 4(𝑥𝑦)2 + 4𝑦 𝑣ậ𝑦 𝑥 = ±2𝑦 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) 2𝑥(4𝑥 − 𝑦 + 2) + 4𝑥 (𝑦 − 1) = √𝑦 − − 2𝑥 + 𝑦 − 3𝑦 + 2(1) 𝑏à𝑖 70: { (√𝑦 − − 1)√2𝑥 + = 8𝑥 − 13(𝑦 − 2) + 82𝑥 − 29(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à: 2𝑥 + ≥ 0, 𝑦 − ≥ 𝑝𝑡(1): 2𝑥(4𝑥 − 𝑦 + 2) + 4𝑥 (𝑦 − 1) = √𝑦 − − 2𝑥 + (𝑦 − 2)(𝑦 − 1) 2𝑥(4𝑥 − 𝑦 + 2) + (𝑦 − 1)(4𝑥 − 𝑦 + 2) = √𝑦 − − 2𝑥 (2𝑥 + 𝑦 − 1)(4𝑥 − 𝑦 + 2) = −(4𝑥 − 𝑦 + 2) √𝑦 − + 2𝑥 𝑣ậ𝑦 4𝑥 − 𝑦 + = ℎ𝑎𝑦 2𝑥 − 𝑦 + = (2𝑥 + 1) + (𝑦 − 2) =− (𝑝𝑡𝑣𝑛 𝑑𝑜 đ𝑘) √𝑦 − + 2𝑥 𝑣ớ𝑖 4𝑥 − 𝑦 + = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑔𝑖ả𝑖 𝑛ó 𝑡ℎ𝑒𝑜 𝑝𝑝 đ𝑜á𝑛 𝑛𝑔ℎ𝑖ệ𝑚 20√6 − 𝑥 − 17√5 − 𝑦 − 3𝑥√6 − 𝑥 + 3𝑦√5 − 𝑦 = 0(1) 𝑏à𝑖 71: { 2√2𝑥 + 𝑦 + + 3√3𝑥 + 2𝑦 + 11 = 𝑥 + 6𝑥 + 13(2) 𝑔𝑖ả𝑖 2𝑥 + 𝑦 + ≥ đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≤ 6, 𝑦 ≤ 𝑣à { 3𝑥 + 2𝑦 + 11 ≥ 𝑝𝑡(1): (20 − 3𝑥)√6 − 𝑥 = (17 − 3𝑦)√5 − 𝑦 2√6 − 𝑥 + 3(√6 − 𝑥) = 2√5 − 𝑦 + 3(√5 − 𝑦) 𝑥é𝑡 𝑓(𝑡) = 2𝑡 + 3𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = + 9𝑡 > 𝑛ê𝑛 − 𝑥 = − 𝑦 𝑣ậ𝑦 𝑥 − 𝑦 = (𝑡ℎế 𝑣à𝑜 (2)) (5 + 16𝑥 −2𝑦 ) 72𝑦−𝑥 +2 = + 16.4𝑥 −2𝑦 (1) 𝑏à𝑖 72: { 𝑥 + 17𝑥 + 10𝑦 + 17 = 2(𝑥 + 4)√4𝑦 + 11(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ − 11 𝑝𝑡(1)đặ𝑡 𝑡 = 𝑥 − 2𝑦 ∶ (5 + 16𝑡 ) 7−𝑡+2 = + 16.4𝑡 𝑣ậ𝑦 𝑡 = => 𝑥 − 2𝑦 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔 12𝑥 − 𝑦(2𝑥 − 4𝑥 + 3) − 12𝑥 + = 0(1) 𝑏à𝑖 73: { 512𝑦 + √ 3√3𝑦 + 𝑦! − 362882 − √3 = 0(2) 𝑔𝑖ả𝑖 để ý 𝑝𝑡(1): (12 − 2𝑦)𝑥 + 𝑥(4𝑦 − 12) − 3𝑦 + = 𝑑𝑒𝑛𝑡𝑎 = 16𝑦 − 96𝑦 + 144 − 4(−3𝑦 + 3)(12 − 2𝑦) = 16𝑦 − 96𝑦 + 144 − 4(−36𝑦 + 6𝑦 + 36 − 6𝑦) 𝑑𝑒𝑛𝑡𝑎 = −8𝑦 + 72𝑦 ≥ 𝑣ậ𝑦 ≤ 𝑦 ≤ 𝑛ℎậ𝑛 𝑡ℎấ𝑦 𝑟ằ𝑛𝑔 𝑦 𝑝𝑡(2): 512 + √ 3√3𝑦 + 𝑦! = 5129 + √ √3.9 + 9! 𝑣ậ𝑦 𝑦 = 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑡ℎế 𝑣à𝑜 𝑡ì𝑚 đượ𝑐 𝑥 = (√𝑥 + − 1) (√𝑦 + + 𝑦) = √𝑥(1) 𝑏à𝑖 74: { 2𝑥 (𝑦 + 1) − (𝑥 + 1)𝑥𝑦 = 2(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ −1 𝑇𝐻1: 𝑛ế𝑢 𝑥 = 𝑡ℎì 𝑝𝑡𝑣𝑛 𝑇𝐻2: 𝑛ế𝑢 𝑥 ≠ 𝑡ℎì 𝑡𝑎 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 √𝑥: 1 √1 + + = √𝑦 + + 𝑦 𝑥 √𝑥 𝑥é𝑡 𝑓(𝑡) = √𝑡 + + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 𝑣ậ𝑦 √𝑥 = 𝑦 => 𝑡 √𝑡 + + > 0( 𝑑𝑜 đ𝑘 𝑙à 𝑡 ≥ −1) = 𝑥(đ𝑘 𝑦 ≠ 0) 𝑦2 2(𝑥 − 2𝑦 − 1)√𝑥 − + 2(𝑥 + 4𝑦 + 2)√𝑥 + = 𝑥 + 24(1) 𝑏à𝑖 75: { √𝑥 − 2𝑦 − + 2√𝑥 + 4𝑦 + = √2𝑦 + + √𝑥 + 28𝑦 + 14(2) 𝑔𝑖ả𝑖 𝑥≥1 𝑥 − 2𝑦 − ≥ 𝑥 + 4𝑦 + ≥ đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦≥− {𝑥 + 28𝑦 + 14 ≥ 𝑝𝑡(2): √𝑥 − (2𝑦 + 1) − √2𝑦 + = √𝑥 + 14(2𝑦 + 1) − √4𝑥 + 8(2𝑦 + 1) 𝑥 − 2(2𝑦 + 1) √𝑥 − (2𝑦 + 1) + √2𝑦 + = −3(𝑥 − 2(2𝑦 + 1)) √𝑥 + 14(2𝑦 + 1) + √4𝑥 + 8(2𝑦 + 1) 𝑣ậ𝑦 𝑥 − 2(2𝑦 + 1) = ℎ𝑎𝑦 𝑝𝑡 𝑐ò𝑛 𝑙ạ𝑖 𝑣ơ 𝑛𝑔ℎ𝑖ệ𝑚 𝑑𝑜 𝑉𝑇 > 𝑚à 𝑉𝑃 < 𝑡ℎế 𝑣à𝑜 𝑝𝑡(1)𝑣ớ𝑖 2𝑦 + = 𝑥 (4𝑥 + 2)√𝑥 + 𝑥 + + (2𝑥 − 𝑦)√𝑦 + 4𝑥 − 4𝑥𝑦 + = 𝑦 − 4𝑥 − 1(1) 𝑏à𝑖 76: { √𝑦 + + √2 − 𝑥 = 𝑥 + 𝑦 − 2(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ −4 𝑣à 𝑥 ≤ 𝑝𝑡(1): (4𝑥 + 2)√𝑥 + 𝑥 + + (2𝑥 − 𝑦)√(2𝑥 − 𝑦)2 + = 𝑦 − 4𝑥 − (2𝑥 + 1)√(2𝑥 + 1)2 + + (2𝑥 + 1) = (𝑦 − 2𝑥)√(𝑦 − 2𝑥)2 + + (𝑦 − 2𝑥) 𝑥é𝑡 𝑓(𝑡) = 𝑡√𝑡 + + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) > 𝑛ê𝑛 𝑡𝑎 𝑐ó 2𝑥 + = 𝑦 − 2𝑥 => 4𝑥 + =𝑦 2𝑥 𝑦 + 3𝑥𝑦 = 4𝑥 + 9𝑦(1) 𝑏à𝑖 77: { 7𝑦 + = 2𝑥 + 9𝑥(2) 𝑔𝑖ả𝑖 4𝑥 2𝑥 + 9𝑥 − 𝑝𝑡(1) 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2): 𝑦 = = 2𝑥 + 3𝑥 − 28𝑥 = (2𝑥 + 3𝑥 − 9)(2𝑥 + 9𝑥 − 6)(∗∗) −9 ± 3√33 𝑝𝑡(∗∗): 𝑥 = −2, 𝑥 = , 𝑥 = (𝑡ℎế 𝑡ừ𝑛𝑔 𝑐á𝑖 𝑣à𝑜 𝑡𝑎 𝑡ì𝑚 đượ𝑐 𝑦) 𝑏à𝑖 78: { 𝑥 + 𝑦 + 𝑧 = 20102 (1) 𝑥 + 𝑦 + 𝑧 = 20103 (2) 𝑔𝑖ả𝑖 𝑡ừ 𝑝𝑡(1): 𝑡𝑎 𝑠𝑢𝑦 𝑟𝑎 đượ𝑐 |𝑥|, |𝑦|, |𝑧| ≤ 2010 𝑠𝑢𝑦 𝑟𝑎 𝑥 + 𝑦 + 𝑧 ≤ |𝑥 | + |𝑦 | + |𝑧 | ≤ 2010(𝑥 + 𝑦 + 𝑧 ) = 20103 𝑥 = 𝑣à 𝑥 = 2010 𝑑ấ𝑢 =𝑥ả𝑦 𝑟𝑎 𝑘ℎ𝑖 {𝑦 = 𝑣à 𝑦 = 2010 𝑧 = 𝑣à 𝑧 = 2010 𝑥 + + 𝑦 + 𝑥𝑦 = 𝑦(1) 𝑦 𝑏à𝑖 79: { 𝑥+𝑦−2= (2) + 𝑥2 𝑔𝑖ả𝑖 𝑣ì 𝑦 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑙à 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑛ê𝑛 𝑡𝑎 𝑥é𝑡 𝑦 ≠ 𝑘ℎ𝑖 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 𝑦 𝑥2 + +𝑦+𝑥 =1 𝑦 𝑦 𝑥2 + đặ𝑡 𝑎 = ,𝑏 = 𝑥 + 𝑦 𝑦 𝑎+𝑏 =1 => 𝑎 = 1, 𝑏 = { 𝑏−2= 𝑎 𝑏à𝑖 80: { √𝑥 − 𝑦 + 3𝑦 𝑥 + √6𝑥 − 𝑥 = 𝑥√2𝑦 − + 𝑥(1) √2𝑥 − − √𝑦 − = 𝑥 − 𝑥 − 𝑦 − 5(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≥ , 𝑦 ≥ 𝑝𝑡(1): 𝑡𝑎 𝑣ì 𝑥 = 𝑘ℎô𝑛𝑔 𝑝ℎả𝑖 𝑛𝑔ℎ𝑖ệ𝑚 𝑐ủ𝑎 ℎ𝑝𝑡 𝑛ê𝑛 𝑥é𝑡 𝑥 ≠ 𝑡𝑎 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 𝑥: 𝑦 𝑦 √1 − ( ) + ( ) + √6𝑥 − = √2𝑦 − + 𝑥 𝑥 𝑦 𝑦 −( ) + 3( ) 𝑥 𝑥 3 𝑦 𝑦 𝑦 𝑦 ( √1 − (𝑥 ) + (𝑥 ) ) + √1 − (𝑥 ) + (𝑥 ) + = 2𝑦 − 6𝑥 √2𝑦 − + √6𝑥 − 𝑦 𝑦 − ( ) ( − 3) 𝑥 𝑥 = 𝑦 𝑦 𝑦 𝑦 ( √1 − (𝑥 ) + (𝑥 ) ) + √1 − (𝑥 ) + (𝑥 ) + 2(𝑦 − 3𝑥) √2𝑥 − + √6𝑥 − 𝑣ậ𝑦 𝑦 = 3𝑥 𝑣à 𝑝𝑡 𝑐ò𝑛 𝑙ạ𝑖 𝑣𝑛 𝑑𝑜 𝑉𝑇 < 𝑚à 𝑉𝑃 >) 𝑡ℎế 𝑦 = 3𝑥 𝑣à𝑜 𝑝𝑡(2) 𝑑à𝑛ℎ 𝑐ℎ𝑜 𝑏ạ𝑛 đọ𝑐 ‼! √𝑥 + 𝑥 + 𝑥 − 𝑦 = 𝑦 − 𝑥(1) 𝑏à𝑖 81: { √𝑥 + 𝑥 − + (𝑦 − 1)√𝑦 + + (𝑥 + 𝑥 − 3)(𝑦 − 1) = 7(2) 𝑔𝑖ả𝑖 𝑝𝑡(1): 𝑥 + 𝑥 + 𝑥 − 𝑦 = 𝑦 − 2𝑥𝑦 + 𝑥 (𝑦 ≥ 𝑥) 𝑥 + 𝑥 − 𝑦 = 𝑦 − 2𝑥𝑦 𝑦 + 𝑦(−2𝑥 + 1) − 𝑥 − 𝑥 = 𝑑𝑒𝑛𝑡𝑎 = 4𝑥 − 4𝑥 + − 4(−𝑥 − 𝑥) = 4𝑥 + + 4𝑥 = (2𝑥 + 1)2 ≥ 𝑦= [ 2𝑥 − − 2𝑥 − = −𝑥 + 𝑥 − 1(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑦 − 𝑥 ≥ 𝑚à − 𝑥 − < 0) 2𝑥 − + 2𝑥 + 𝑦= = 𝑥 + 𝑥(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) 3√𝑥 − 2𝑥𝑦 + 𝑦 + 𝑥 − 𝑦 = 4(1) 𝑏à𝑖 82: { 𝑥√𝑥 − 𝑦 − (𝑥 + 1)√𝑦 + (𝑥 + 2)√𝑦 + = 17(2) 𝑔𝑖ả𝑖 𝑦≥0 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: {𝑦 ≥ −3 𝑥≥𝑦 𝑝𝑡(1): 3√(𝑥 − 𝑦)2 + 𝑥 − 𝑦 = 𝑣ậ𝑦 𝑥 − 𝑦 = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)𝑔𝑖ả𝑖 𝑏ì𝑛ℎ 𝑡ℎườ𝑛𝑔 ‼! 4(𝑥 + 𝑦)√𝑥 − + 4(𝑥 + 3𝑦) 3√𝑥 − 8𝑦 = 3(1) 𝑏à𝑖 83: { 𝑥 + 2𝑦 3 2(𝑥 + 𝑦)√ = √𝑥 + 3𝑥 𝑦 + 𝑥 + 3𝑦(2) 𝑥 𝑔𝑖ả𝑖 𝑥 ≥ 8𝑦 𝑥≥1 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 + 2𝑦 ≥0 𝑥 { 𝑥≠0 𝑝𝑡(2): √𝑥(𝑥 + 2𝑦) + (√𝑥 + 2𝑦) 3 = √𝑥 (𝑥 + 3𝑦) + 𝑥 + 3𝑦 √𝑥 𝑏3 đặ𝑡 𝑎 = √𝑥 + 2𝑦, = 𝑥 + 3𝑦 𝑥 𝑎3 𝑏3 𝑎 √𝑥 + =𝑏+ 𝑥 √𝑥 (𝑎√𝑥 − 𝑏) + (𝑎√𝑥 − 𝑏) + √𝑥 √𝑥 (𝑎 − (𝑎 − ( 𝑏 √𝑥 𝑎2 + 𝑣ậ𝑦 𝑎√𝑥 − 𝑏 = ℎ𝑎𝑦 + 𝑏 √𝑥 ) (𝑎 + ) )=0 𝑎𝑏 √𝑥 + 𝑏2 )=0 𝑥 𝑎𝑏 𝑏 + √𝑥 𝑥 = 0(𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > ) 𝑥 𝑣ậ𝑦 √(𝑥 + 2𝑦)𝑥 = √𝑥 (𝑥 + 3𝑦) (𝑥 + 2𝑦𝑥)3 = 𝑥 (𝑥 + 3𝑦)2 𝑥 + 6𝑥 𝑦 + 12𝑥 𝑦 + 8(𝑥𝑦)3 = 𝑥 (𝑥 + 6𝑥𝑦 + 9𝑦 ) = 𝑥 + 6𝑥 𝑦 + 9𝑥 𝑦 3𝑥 𝑦 + 8(𝑥𝑦)3 = 𝑣ậ𝑦 [ 𝑦=0 3𝑥 + 8𝑦 = (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) (3𝑥 + 3𝑥𝑦) (1 + √1 + 9𝑥 ) (√5 + 𝑦 − √1 + 𝑦) = 8(1) 𝑏à𝑖 84: { −243𝑥 (1 + 𝑦) + 18𝑥 (1 + 𝑦) + 78𝑥 = √27𝑥 − 14(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑦 ≥ −1 để ý 𝑝𝑡(1): 3𝑥(1 + 𝑦) (1 + √1 + 9𝑥 ) = 2(√5 + 𝑦 + √1 + 𝑦) 𝑇𝐻1: + 𝑦 = 𝑣ậ𝑦 𝑦 = −1 𝑝𝑡(1)𝑡ℎấ𝑦 𝑣ô 𝑛𝑔ℎ𝑖ệ𝑚 𝑇𝐻2: + 𝑦 ≠ 𝑡𝑎 𝑐ℎ𝑖𝑎 𝑣ế 𝑐ủ𝑎 𝑝𝑡(1)𝑐ℎ𝑜 + 𝑦 5+𝑦 3𝑥 (1 + √1 + (3𝑥)2 ) = 2√ + (1 + 𝑦)2 √1 + 𝑦 3𝑥 + √(3𝑥)2 𝑥é𝑡 𝑓(𝑡) = 𝑡 + + (3𝑥)4 √𝑡 𝑡4 + ′ (𝑡) =1+ 𝑣ậ𝑦 𝑓(𝑡)đơ𝑛 đ𝑖ệ𝑢 𝑡ă𝑛𝑔 𝑛ê𝑛 ∶ 3𝑥 = = √( ) +( ) +( ) √1 + 𝑦 √1 + 𝑦 √1 + 𝑦 𝑣ậ𝑦 𝑓 2𝑡 + 4𝑡 2√𝑡 + 𝑡 √1 + 𝑦 =1+ + 2𝑡 √𝑡 + => + 𝑦 = 4𝑥𝑦 + = 𝑥 + 2√𝑥𝑦(1) 𝑏à𝑖 85: { −1 −1 (𝑥 √𝑥) + 8𝑦√𝑦 = (√𝑥) + 6√𝑦(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 > 𝑣à 𝑦 ≥ 9𝑥 >0 𝑝𝑡(2): 1 + 𝑦) = + √𝑦 (2 √ 𝑥 √ (√𝑥) 3 + 2(√𝑥𝑦) = 𝑥 + 6(√𝑥) √𝑦 𝑙ấ𝑦 𝑝𝑡(1)𝑡ℎế 𝑝𝑡(2): 2√𝑥𝑦 − 4𝑥𝑦 + 2(√𝑥𝑦) = 6𝑥(√𝑥𝑦) 𝑣ậ𝑦 𝑥 = ℎ𝑎𝑦 𝑦 = ℎ𝑎𝑦 − 4√𝑥𝑦 + 2𝑥𝑦 = 6𝑥 𝑣ớ𝑖 𝑥 = − 4√𝑥𝑦 + 2𝑥𝑦 (𝑡ℎế 𝑣à𝑜 𝑝𝑡(1) (2015 − 2𝑥)√4 − 𝑥 + (4𝑦 − 2013)√3 − 2𝑦 = 0(1) 𝑏à𝑖 86: { 2√7𝑥 − 8𝑦 + 3√14𝑥 − 18𝑦 = 𝑥 + 6𝑥 + 13(2) 𝑔𝑖ả𝑖 𝑥≤4 𝑦≤ đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 7𝑥 − 8𝑦 ≥ {14𝑥 − 18𝑦 ≥ 𝑝𝑡(1): 2007(√4 − 𝑥) + 2(√4 − 𝑥) = 2007(√3 − 2𝑦) + 2(√3 − 2𝑦) 𝑥é𝑡 𝑓(𝑡) = 2007𝑡 + 2𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 2007 + 3𝑡 > 𝑛ê𝑛 − 𝑥 = − 2𝑦 => 𝑥 − 2𝑦 = (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) 𝑥 − (𝑥𝑦 + ) = √3𝑥 + + 3(1) 𝑥 𝑏à𝑖 87: (đ𝑘 𝑥, 𝑦 12𝑥 + 8𝑥 2 (2𝑥 + 1) − (2) = (2𝑥𝑦 + 1) + 8𝑦 − 𝑥2 𝑥 𝑦+1 { ∈ [−1; +∞)) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 ≠ 𝑝𝑡(2): 4𝑥 + 4𝑥 − 12 − = 4(𝑥𝑦)2 + 4𝑥𝑦 + 8𝑦 − 𝑥 𝑥 𝑥𝑦 + 𝑥 8 4𝑥 − + 4𝑥 = (𝑥𝑦 + ) − + (𝑥𝑦 + ) 𝑥 𝑥 𝑥 𝑥𝑦 + 𝑦 𝑥é𝑡 𝑓(𝑡) = 4𝑡 − + 4𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) > 𝑡 𝑣ậ𝑦 𝑥 = 𝑥𝑦 + 𝑏à𝑖 88: { (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑑ù𝑛𝑔 𝑐𝑎𝑟𝑛𝑎𝑑𝑜 𝑐ℎ𝑜 𝑝𝑡 𝑏ậ𝑐 3) 𝑥 2017𝑥−𝑦−2 (√𝑥 + + 𝑥) (√𝑦 + 4𝑦 + − 𝑦 − 2) = 1(1) 2√𝑦 + 4𝑦 + = 2√𝑥 − + 𝑥 (2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ 𝑥 ≥ 𝑝𝑡(1): 2017𝑥 (√𝑥 + + 𝑥) = (√(𝑦 + 2)2 + + (𝑦 + 2)) 2017𝑦+2 𝑡𝑎 𝑥é𝑡 𝑓(𝑡) = 2017𝑡 (√𝑡 + + 𝑡) 𝑣ậ𝑦 𝑓 ′ (𝑡) > 𝑣ậ𝑦 𝑥 = 𝑦 + (𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑡𝑎 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 𝑙à 𝑦 = 0, 𝑥 = 2) (3𝑦 + 8)√𝑥 + = 10𝑦 − 3𝑥𝑦 + 12(1) 𝑏à𝑖 89: { (𝑏𝑦 𝑁𝑔𝑢𝑦ễ𝑛 𝑇ℎế 𝑇â𝑛 ) 5𝑦 √2 − 𝑥 − = 6𝑦 + 𝑥𝑦 √2 − 𝑥(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚 ∶ −2 ≤ 𝑥 ≤ 𝑝𝑡(2): (5 − 𝑥)𝑦 √2 − 𝑥 − = 6𝑦 (𝑦√2 − 𝑥) − + 3𝑦 √2 − 𝑥 − 6𝑦 = (𝑦√2 − 𝑥 − 2)(𝑦 (2 − 𝑥) + 2𝑦√2 − 𝑥 + 4) + 3𝑦 (𝑦√2 − 𝑥 − 2) = 𝑣ậ𝑦 𝑦√2 − 𝑥 − = ℎ𝑎𝑦 𝑦 (2 − 𝑥) + 2𝑦√2 − 𝑥 + + 3𝑦 = (𝑝𝑡𝑣𝑛 𝑑𝑜 𝑉𝑇 > 0) 𝑣ớ𝑖 𝑦√2 − 𝑥 − = 𝑡ℎế 𝑣à𝑜 𝑝𝑡(2) 𝑡𝑎 đượ𝑐 𝑛𝑔ℎ𝑖ệ𝑚 𝑥 = 1.2 √𝑥 − 𝑦 + + √3𝑥 + 2𝑦 + = 3𝑥 + 1(1) (𝑏𝑦 𝑁𝑔𝑢𝑦ễ𝑛 𝑇ℎế 𝑇â𝑛) 𝑏à𝑖 90: { 𝑥√𝑥 − + √𝑥 + 3𝑦 + = (𝑦 + 5)√𝑦 + 1(2) 𝑔𝑖ả𝑖 𝑦≥0 𝑥≥2 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑥 − 𝑦 + ≥ 3𝑥 + 2𝑦 + ≥ { 𝑥 + 3𝑦 + ≥ 3 𝑝𝑡(2): (√𝑥 − 2) − (√𝑦 + 1) + 2√𝑥 − − 4√𝑦 + + √𝑥 + 3𝑦 + = 2 (√𝑥 − − √𝑦 + 1) ((√𝑥 − 2) + √(𝑥 − 2)(𝑦 + 1) + (√𝑦 + 1) ) + 2(√𝑥 − − √𝑦 + 1) − √4𝑦 + + √𝑥 + 3𝑦 + = 2 (𝑥 − 𝑦 − 3) ((√𝑥 − 2) + √(𝑥 − 2)(𝑦 + 1) + (√𝑦 + 1) ) + (√𝑥 − + √𝑦 + 1) 𝑥−𝑦−3 √𝑥 + 3𝑦 + + √4𝑦 + + 2(𝑥 − 𝑦 − 3) √ 𝑥 − + √𝑦 + =0 𝑥 − 𝑦 − = 0(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) 𝑣ậ𝑦 [ (√𝑥 − 2) + √(𝑥 − 2)(𝑦 + 1) + (√𝑦 + 1) (√𝑥 − + √𝑦 + 1) + + (√𝑥 − + √𝑦 + 1) (√𝑥 + 3𝑦 + + √ 𝑥 + 3𝑥 + 𝑦 2𝑦 − 𝑥 − √ − √𝑦 − = (1) 2 𝑏à𝑖 91: 3 2 {𝑥 − 5𝑥 − 4𝑦 + 32𝑦 + √𝑥 − 35 = √2(𝑦 − 1) + √−𝑥 + 9𝑥 − 19𝑥 + 11(2) 𝑔𝑖ả𝑖 đ𝑘 𝑐ó 𝑛𝑔ℎ𝑖ệ𝑚: 𝑦 ≥ 1(∗) 𝑝𝑡(1): √( 𝑥+2 𝑥+2 𝑥+2 𝑦 ) −( ) + ( ) = √𝑦 − + 𝑦 2 2 2𝑡 − 𝑡 + > 0(𝑑𝑜 đ𝑘(∗) => 𝑡 ≥ 1) 𝑥é𝑡 𝑓(𝑡) = √𝑡 − + 𝑡 𝑣ậ𝑦 𝑓 ′ (𝑡) = 𝑡 2√ 𝑡 − 𝑣ậ𝑦 𝑥+2 = 𝑦(𝑡ℎế 𝑣à𝑜 𝑝𝑡(2)) THÔI !!CHUYÊN ĐỀ VỀ HPT CỦA CHÚNG TA KẾT THÚC TẠI ĐÂY ĐƯỢC RỒI !! BỞI VÌ KHƠNG CĨ TIME +PHIÊN BẢN NÊN KHÔNG THỂ KHÔNG TRÁNH KHỎI SAI SĨT VÀ THIẾU SĨT MONG AE THƠNG CẢM BỎ QUA Ạ !!!RẤT VUI LỊNG VỚI SỰ GĨP Ý CỦA AE CHÚC CÁC BẠN ĐỌC ĐƯỢC NHIỀU SỨC KHỎE VÀ THÀNH CÔNG TRONG CUỘC SỐNG NÀY NHA !!!!