hướng dẩn tài liệu ôn ds9 chương1,2

Hớng dẫn tập đáp án đề kiểm tra Chơng i Căn bậc hai bậc ba A- §èi víi häc sinh TB – Ỹu Bµi 1: a) x ≥ − b) Víi mäi x ; Bµi 2: a) < Bµi 3: b) > 17 c) x ≤ 1 > 2 c) a) 12 75 = 4.3.3.25 = 2.3.5 = 30 b) 24 36 = 25 49 36 = = 14 c) 25 25 21 = 12 25 25 21 = 12 d) 0, 04.25 = 0,2 = f) 35 5 25 = e) 90.6, = 24 25 25 = = = 16 16 16 25 25 = = 121 121 11 h) 25 2 1 = = = = 18 18 Bµi 4: a x = ± Bµi 5: a) g) b x = ± 10 + 20 + 80 = c x = ± 5+2 5+4 =7 b) + 12 + 24 = c) x − 4x + 16x = Bµi 6: a) x = b) x = + + 12 = 15 x − x + x = x (x ≥ ) Bµi 7: a) (1 − ) = − = − b) ( − 2)2 + = − + = − + = Bµi 8: a) (−7) + = + = 14 b) ( − 3) + (2 − ) = −3 + 2− = 3− + −2 =1 Bµi 10: a) −1 −1 = = 5− 5 ( − 1) Bµi 11: a) 1 + 2− 2+ b) c) Bµi 1− − 1+ = = b) a +1 = a +a a c) a −1 = a −1 a +1 2+ 2− + =4 (2 − )(2 + ) ( + )(2 − ) 2+2 −2 − = (1 − )(1 + ) (1 + )(1 − ) +2− +2 = −4 1− 1 2+ 2+ + = + = 2+ 3−2− = 3− 4−3 4−5 2− 2− 1 x +1 x −1 + = + = x −1 x + ( x − 1)( x + 1) ( x + 1)( x − 1) 12:a) x + 1+ x −1 x = x −1 x −1 b) x− y x+ y 1 − = − = x+ y x − y ( x + y )( x − y ) ( x − y )( x + y ) 1  x −9 − =  x +3  x −3 Bµi 13 : a)   x− y− x− y x = x− y x− y x +3− x +3 x −9 =3 x −9 b)  x −2  ( x − 2) + ( x + 2)  x + 2  : .( x − 4) = x − x + + x + x + = x + + =  x+2  x−4 x −2 x−4      B- Đối với học sinh Khá, giỏi Bài 1: a) §Ĩ 2− x −4 cã nghÜa th× - x2 − ≠ ⇔ ≤ x - ≠ ⇔ x ≥ hc x ≤ -2 1 b) §Ĩ − x + cã nghÜa th× - 9x2 ≥ ⇔ − ≤ x ≤ 3 c) §Ĩ 2x − x − 2x − th× 1    2 x − ≥ x ≥ x ≥ x ≥ 2 ⇔ ⇔ ⇔ ⇔ ≤ x ≠1  x − 2x − > 2 x − − 2 x − + > ( x − − 1) > 2 x − ≠    ;  x ≥ 2x − ≥  x − + 3x th×  ⇔ ⇔ x≥  2x + 2 x + ≠ x ≠ − d) Để Bài 2: a, b, c, a2 a −1 −5 a+6 cã nghÜa a ≠ cã nghÜa a – > ⇒ a > d, a + cã nghÜa a < - cã nghÜa víi mäi a ( 5) Bµi 3: a) + = ( 3) b) 5−2 = c) − 10 + = 2 + + 12 = − + ( 2) ( + 2= 5− ) 2 = ( ) +1 ( = 3− ) +1 = +1 = 3− = 3− 5− + = 5− 2+ = d) 8a − 5a = (3 2a ) − 5a = 2a − 5a = −3a Bµi 4: A = (2 + ): - 2 = + − 2 = − B = 14 − − 14 + = (3 − )2 − (3 + ) = −2 ( Có thể làm theo cách C2 = = 20, C < nªn C = 20 ) C = - ( HD: C¸ch 1: B = 2; C¸ch 2: B2 = B < nªn B = - ) D= − − (3 − 5) = − − = − (1 − ) = − + = E= − − 21 − 12 =1 Bµi 5: a) 16 x = ⇔ 42 x = ⇔ x = ⇔ x = ⇔ x = 5 b) x = ⇔ 22 x = ⇔ x = ⇔ x = ⇔x = c) ( x − 1) = 21 ⇔ x − = 21 ⇔ x − = ⇔ x − = 49 ⇔ x = 50 d) ( − x ) − = ⇔ ( − x ) − = ⇔ ( 1− x) = 3⇔ 1− x = 3; 1 − x =  x = −2 ⇔ ⇔ 1 − x = −3  x = Bµi : A = x − x + = − x + x − 3x 3x + B= 1 a = − + a − 2 a + a −1 a −1 C=  x +1 : + x − x +1  x  − x+ x x −1 x x−2  =  x −1 x− x Bµi A = x + x − + x − x − Ta cã A = 2x + 2 x - + x - 2x - = 2x − + 2x - + + x − - 2x - + = ( 2x - + 1)2 + ( 2x - − 1)2 Do A = ( 2x - + 1)2 + ( 2x - 1)2 điều kiện xác định A lµ 2x – ≥ ⇔ x ≥ Khi 2x − + + ®ã A   x − + − 2x − + = nÕu ≤ x < 2x − − =   x − + + 2x − − = 2x − nÕu ≤ x    nÕu ≤ x < ⇒A=   x − nÕu ≤ x  B = x + 2 x − + x − 2 x − = ( x − + )2 + ( x − − ) §KX§ : x ≥ B = 2 nÕu ≤ x x −4 x −4 −2 >1⇔ −1 > ⇔ >0 ⇔ x −21 by Bµi 11; Ta cã ax = by ⇒ ax = x 3 T¬ng tù ta cã: by2 = cz ; cz2 = ax y z 3 cz Suy ra: ax2 + by2 + cz2 = by + + ax x y z x +3 x +3 x −1 c) A = ) ( x +3 − x −1 ( 2) = (3− 2) −2 = = −( + ) 2− 2 ) ) ) x +3 ) ) x −1 = −2 x −1 1 Mµ ax3 = by3 = cz3 ⇒ ax2 + by2 + cz2 = ax3( x + y + z ) = ax3 = by3 = cz3 ax + by + cz ax + by + cz ax + by + cz ⇒a= ;b= ;c= y3 x3 z3 a +3 b +3 c = VËy 3 ax + by + cz ( 1 + + ) = ax + by + cz x y z ax + by + cz = a + b + c x ≥ x Bài 12: a) Để P có nghÜa th×  b) Ta cã P = x ( x − 3) − ( x − 1)( x + 3) − − 11 x x−9 x−9 x−9 = 2x − x − x − x + − + 11 x = x + x = x−9 x−9 x x −3  x  x 