Question

Chứng tỏ bt sau ko phụ thuộc vào biến:

C= (x-1)^3 - (x+1)^3 + 6(x+1) (x-1)

Answer 1

\(C=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(C=x^3-2x^2+x-x^2+2x-1-x\left(x^2+2x+1\right)-x^2-2x-1+6x^2-6x+6x-6\)

\(C=x^3-2x^2+x-x^2+2x-1-x\left(x^2+2x+1\right)-x^2-2x-1+6x^2-6\)

\(C=x^3+2x^2+x-8-x\left(x^2+2x+1\right)\)

\(C=x^3+2x^2+x-8-x^3-2x^2-x\)

\(C=-8\left(đpcm\right)\)

Answer 2

C = (x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

C = x³ - 3x² + 3x - 1 - x³ - 3x² - 3x - 1 + 6(x + 1)(x - 1)

C = x³ - 3x² + 3x - 1 - x³ - 3x² - 3x - 1 + 6x² - 6

C = (x³ - x³) + (-3x² - 3x² + 6x²) + (3x - 3x) + (-1 - 1 - 6)

C = -8

Vậy: biểu thức không phụ thuộc vào biến

Answer 3

\(C=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(C=x^3-3x^2+3x-1-\left(x^3+3x^2+3x+1\right)+6\left(x^2-1\right)\)

\(C=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

\(C=-8\)

Vậy bt  trên ko phụ thuộc vào biến.

Answer 4

\(C=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right).\)

\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]\)\(+6\left(x^2-1\right)\)

\(=-2\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)

\(=-2\left(x^3-1\right)+6x^2-6=-6x^3+2+6x^2-6=-4\)