Question

[Minh Tran Tú]

Rút gọn biểu thức.

a) \(\frac{1}{2}x\left(1+2x\right)+\left(1-x\right)\left(x+2\right)\)

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

c) \(x\left(x-1\right)\left(x+1\right)-\left(x+1\right)\left(x^2-x+1\right)\)

Answer 1

[Akai Haruma]

Lời giải:

a)

$\frac{1}{2}x(1+2x)+(1-x)(x+2)$

$=\frac{1}{2}x+x^2+x+2-x^2-2x$

$=\frac{-1}{2}x+2$

b)

$(2x-1)^3-(3+2x)(2x-3)+8x^2(2-x)$

$=(8x^3-12x^2+6x-1)-(4x^2-9)+(16x^2-8x^3)$

$=6x+8$

c)

$x(x-1)(x+1)-(x+1)(x^2-x+1)$

$=(x^2-x)(x+1)-(x+1)(x^2-x+1)$

$=(x+1)[(x^2-x)-(x^2-x+1)]=(x+1)(-1)=-(x+1)$