Toán Lớp 8: Tìm x, biết: a) (2x-1) ³ +4(x-1)(x+1-2x ²)=7 b) (x+1) ³ + (x-2) ³-2x²(x-1,5)=3 c)(x +2 )(x ² – 2x + 4) – x (x –3)(x + 3) = 26. d) x ² –

Question

Toán Lớp 8:  Tìm x, biết: a) (2x-1) ³ +4(x-1)(x+1-2x ²)=7 b) (x+1) ³ + (x-2) ³-2x²(x-1,5)=3 c)(x +2 )(x ² – 2x + 4) – x (x –3)(x + 3) = 26. d) x ² – ( 2x -1) ² = 0

phuongthuydung · Answer

a)   (2x-1)^3 + 4 (x-1)(x+1-2x^2) = 7   => [ (2x)^3-  3 . (2x)^2 . 1 + 3 . 2x . 1^2 - 1^3] + (4x-4)(x+1-2x^2) =7   => (8x^3-  12x^2 + 6x-1) + (4x^2 +4x - 8x^3 - 4x - 4 + 8x^2) = 7   => 8x^3-  12x^2 + 6x-1 + 4x^2 +4x - 8x^3 - 4x - 4 + 8x^2 = 7   => (8x^3 - 8x^3) + (4x^2 + 8x^2 - 12x^2) + (6x+ 4x - 4x) - 1 - 4 =7   => 6x-5=7   => 6x = 12   =>x=2   Vậy x=2   b)   (x+1)^3 + (x-2)^3- 2x^2( x-1,5) =3   => (x^3 + 3 . x^2 . 1 + 3 . x . 1^2 + 1^3) + (x^3 - 3 . x^2 . 2 + 3 . x.  2^2 - 2^3) - (2x^2 . x - 2x^2 . 1,5) = 3   => (x^3 + 3x^2 + 3x+1) +(x^3 - 6x^2 +12x-8) - (2x^3 - 3x^2) = 3   => x^3 + 3x^2 + 3x + 1 + x^3 - 6x^2 + 12x - 8 - 2x^3 + 3x^2 = 3   => (x^3 + x^3 - 2x^3) + (3x^2 - 6x^2 + 3x^2) + (3x + 12x) + (1-8) =3   => 15x - 7 =3   => 15x=10   => x=2/3   Vậy x=2/3   c)   (x+2)(x^2 - 2x+4) - x (x-3)(x+3) =26   => (x+2)(x^2-  x.  2 + 2^2) - x (x^2 -3^2) =26   => (x^3 + 2^3) - x (x^2 - 9) =26   => x^3 + 8 - x^3+  9x = 26   => 9x + 8 = 26   => 9x=  18   =>x=2   Vậy x=2   d)   x^2 - (2x-1)^2 =0   => [ x - (2x-1)] [ x + (2x-1)] = 0   => (x-2x+1)(x+2x-1)=0   => (1-x)(3x-1)=0   =>1-x=0 hoặc 3x-1=0   +)1-x=0=>x=1   +)3x-1=0=>3x=1=>x=1/3   Vậy x  in {1;1/3}