Ta có : \(\left(x-2\right)^2-\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(x-2-x-3\right)\)

\(=-5\left(x-2\right)\)

Ta có : \(\left(x+1\right)^2+\left(x-3\right)^2-2\left(x+1\right)\left(x-3\right)\)

\(=\left(x+1-x+3\right)^2=4^2=16\)

`(x-2)^2-(x-2)(x+3)`

`=x^2-4x+4-(x^2+3x-2x-6)`

`=x^2-4x+4-x^2-x+6`

`=-5x+10`

`b)(x+1)^2+(x-3)^2-2(x+1)(x-3)`

`=(x+1)^2-2(x+1)(x-3)+(x-3)^2`

`=[(x+1)-(x-3)]^2`

`=(x+1-x+3)^2`

`=4^2=16`

Lời giải:
ĐKXĐ: $x\neq -1; x\neq 0; x\neq 2$

\(Q=1+\left[\frac{x+1}{(x+1)(x^2-x+1)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right]:\frac{x^2(x-2)}{x(x^2-x+1)}\)

\(=1+\left[\frac{1}{x^2-x+1}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right].\frac{x^2-x+1}{x-2}\)

\(=1+(\frac{2}{x^2-x+1}-\frac{2}{x+1}).\frac{x^2-x+1}{x-2}\\ =1+\frac{2}{x-2}-\frac{2(x^2-x+1)}{(x+1)(x-2)}=\frac{x}{x-2}-\frac{2x^2-2x+2}{(x+1)(x-2)}\)

\(=\frac{x(x+1)-(2x^2-2x+2)}{(x+1)(x-2)}=\frac{-x^2+3x-2}{(x+1)(x-2)}=\frac{(1-x)(x-2)}{(x+1)(x-2)}=\frac{1-x}{1+x}\)

a)(x + 1)² – (x – 2)²

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

= 3(2x-1)

b)(x – 3)(x – 1) – (2x – 1)²

= x²-4x+3-4x²+4x-1

= -(3x²-2)

c)(x + 3)² - 2(x + 3)(1 – x) + (1 - x)²

= [(x+3)-(1-x)]²

=(2x-2)²=4(x-1)²

Làm như vậy nè :

=X-1/X-2 : (X-2/X-3 x X-1/X-3)

=X-1/X-2 : (X-2)x(X-1)/(X-3)² 

=X-1/X-2 x (X-3)² /(X-2) x(X-1)

=(x-3)²/(X-2)²

DẤU / LÀ DẤU PHẦN NHÉ

\(2x\left(x-2\right)+5\left(x+3\right)+3\left(x+1\right).\)

\(=2x^2-4x+5x+15+3x+3=2x^2+4x+18.\)

\(5x^2-2\left(x+1\right)+3x\left(x-2\right)+5.\)

\(=5x^2-2x-2+3x^2-6x+5=8x^2-8x+3.\)

\(a/4x\left(x-3\right)-3x\left(2+x\right)\\ =4x.x-4x.3-3x.2-3x.x\\ =4x^2-12x-6x-3x^2\\ =x^2-18x\\ b/2x\left(5x+2\right)+\left(2x-3\right)\left(3x-1\right)\\ =2x.5x+2x.2+2x.3x-2x.1-3.3x+3.1\\ =10x^2+4x+6x^2-2x-9x+3\\ =16x^2-7x+3\)

Ta có: \(\left(\dfrac{3}{x-1}+\dfrac{x+3}{x^2-1}\right):\left(\dfrac{x+2}{x^2+x-2}-\dfrac{x}{x+2}\right)\)

\(=\dfrac{3\left(x+1\right)+x+3}{\left(x-1\right)\left(x+1\right)}:\dfrac{x+2-x\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{3x+3+x+3}{\left(x-1\right)\left(x+1\right)}:\dfrac{x+2-x^2+x}{\left(x+2\right)\left(x-1\right)}\)

\(=\dfrac{4x+6}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+2\right)\left(x-1\right)}{-x^2+2x+2}\)

\(=\dfrac{\left(4x+6\right)\left(x+2\right)}{\left(-x^2+2x+2\right)\left(x+1\right)}\)