a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)

\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

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

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

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

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

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

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

x(x+4)(x-4)-(x^2+1)(x^2-1)=x(x²-16)-(x⁴-1)

=x³-16x-x⁴+1

\(T=\dfrac{2\left(x-1\right)}{\sqrt{x}+1}+\dfrac{x-4}{\sqrt{x}-2}\)

\(T=\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)

\(T=2\left(\sqrt{x}+1\right)+\left(\sqrt{x}+2\right)\)

\(T=2\sqrt{x}+2+\sqrt{x}+2\)

\(T=3\sqrt{x}+4\)

\(x=4\)

\(\Rightarrow T=3\sqrt{4}+4=3.2+4=10\)

Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

(x + 1)^4 - 6(x + 1)^2 - (x^2 - 2)(x^2 + 2)

= (x^2 + 2x + 1)(x^2 + 2x + 1) - 6(x^2 + 2x + 1) - (x^2 - 2)(x^2 + 2)

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

= x^4 + 2x^3 + x^2 + 2x^3 + 4x^2 + 2x + x^2 + 2x + 1 - 6x^2 - 12x - 6 - x^2 + 2^2

= 4x^3 - 8x - 1

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

\(=\left(x^2+2x-5\right)\left(x^2+2x+1\right)-x^4+2\)

\(=x^4+2x^3+x^2+2x^3+4x^2+2x-5x^2-10x-5-x^4+4\)

\(=4x^3-8x-1\)

Bro ok

Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x^2-1\right)-x+2\)

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

\(=-6\)

có nhất thiết phải làm ko bạn

a: Ta có: \(\left(x+3\right)^2-\left(x+1\right)\left(x-4\right)\)

\(=x^2+6x+9-x^2+3x+4\)

\(=9x+13\)

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

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

\(=x^3-4x-x^3+2x^2-x\)

\(=2x^2-5x\)

c) (x+3)²-(x+1)(x-4)

=x²+6x+9-(x²-3x+4)

=x²+6x+9-x²+3x-4

=9x+5

d)x(x+2)(x-2)-x(x-1)²

=x(x²-4)-x(x²-2x+1)

=x(x²-4-x²+2x-1)

=x(2x-5)

=2x²-5x