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

Đợi anh chút

\(\dfrac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}=\dfrac{y\left(x^2+2xy+y^2\right)}{2x^2+2xy-xy-y^2}=\dfrac{y\left(x+y\right)^2}{2x\left(x+y\right)-y\left(x+y\right)}\)

\(=\dfrac{y\left(x+y\right)^2}{\left(2x-y\right)\left(x+y\right)}=\dfrac{y\left(x+y\right)}{2x-y}\)

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

a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)

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

\(=\dfrac{x+1}{2x}\)

b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)

\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(8-x\right)}{x+2}\)

\(=\dfrac{x-8}{x+2}\)

c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)

\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)

\(=-2\)

d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

ĐKXĐ: \(x\neq0;x\neq-1\)

\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)

$---$

ĐKXĐ: \(x\neq y\)

\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}x^2-9\ne0\\x+3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne-3\end{matrix}\right.\Leftrightarrow x\ne\pm3\) 

b) \(A=\dfrac{x+15}{x^2-9}-\dfrac{2}{x+3}\)

\(A=\dfrac{x+15}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x+15-2x+6}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{21-x}{\left(x+3\right)\left(x-3\right)}\)

c) Thay x = - 1 vào A ta có: 

\(A=\dfrac{21-\left(-1\right)}{\left(-1+3\right)\left(-1-3\right)}=\dfrac{21+1}{2\cdot-4}=\dfrac{22}{-8}=-\dfrac{11}{4}\)

\(\dfrac{x}{x+2}+\dfrac{7x-16}{\left(x+2\right)\left(4x-7\right)}\left(dkxd:x\ne-2;x\ne\dfrac{7}{4}\right)\)

\(=\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x^2-4\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}\)

\(=\dfrac{4\left(x-2\right)}{4x-7}\)

\(=\dfrac{4x-8}{4x-7}\)

===============================

\(\dfrac{x}{x-2}+\dfrac{2}{2-x}\left(dkxd:x\ne2\right)\)

\(=\dfrac{x}{x-2}-\dfrac{2}{x-2}\)

\(=\dfrac{x-2}{x-2}\)

\(=1\)

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

Để biểu thức trên nhận giá trị âm khi \(\dfrac{\left(x-2\right)^2}{x^3-2x^2-4x+8}< 0\)

\(\Rightarrow x^3-2x^2-4x+8< 0\)do \(\left(x-2\right)^2\ge0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-2x\left(x+2\right)< 0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)^2< 0\Leftrightarrow x< -2\)

ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)

Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)

a) \(P=\dfrac{x^2+3x}{x^2-8x+16}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\right)\left(x\ne0,x\ne4\right)\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x\left(x-4\right)}\right)\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{\left(x+4\right)\left(x-4\right)+x+19-x^2}{x\left(x-4\right)}\)

\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{x+3}{x\left(x-4\right)}=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\dfrac{x\left(x-4\right)}{x+3}=\dfrac{x^2}{x-4}\)

b) \(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\sqrt{3}+1-\sqrt{3}+1=2\)

\(\Rightarrow P=\dfrac{2^2}{2-4}=-2\)

a)\(ĐKXĐ:\left\{{}\begin{matrix}x\left(x-4\right)\ne0\\\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne0\\x\ne-3\end{matrix}\right.\)

\(P=\dfrac{x\left(x+3\right)}{\left(x-4\right)}:\left(\dfrac{x^2-16+x+19-x^2}{x\left(x-4\right)}\right)=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\left(\dfrac{x\left(x-4\right)}{x+3}\right)=\dfrac{x^2}{x-4}\)

b)\(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3+1}-\left(\sqrt{3}-1\right)=2\)

thay x=2 vào P ta có \(P=\dfrac{2^2}{2-4}=-2\)

a: |x-1|=3

=>x-1=3 hoặc x-1=-3

=>x=-2(nhận) hoặc x=4(loại)

Khi x=-2 thì \(A=\dfrac{4+4}{-2-4}=\dfrac{8}{-6}=\dfrac{-4}{3}\)

b: ĐKXĐ: x<>4; x<>-4

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

\(=\dfrac{-x^2-8x-16+x^2-8x+16-4x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{-4x^2-16x}{\left(x-4\right)\left(x+4\right)}\)

=-4x/x-4

c: A+B

=-4x/x-4+x^2+4/x-4

=(x-2)^2/(x-4)
A+B>0

=>x-4>0

=>x>4