a) \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)

\(A=\left(\frac{x}{\left(x-2\right)\cdot\left(x+2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\right)\div\left(1-\frac{x}{x+2}\right)\)

\(A=\frac{x+x-2-2\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{x+2-x}{x+2}\)

\(A=\frac{2x-2-2x-4}{\left(x-2\right)\cdot\left(x+2\right)}\div\frac{2}{x+2}\)

\(A=\frac{-6}{\left(x-2\right)\cdot\left(x+2\right)}\cdot\frac{x+2}{2}\)

\(\Rightarrow A=\frac{-3}{x-2}\)

b) Với x = -4 . Ta có : 

\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{-3}{-6}=\frac{1}{2}\)

cho tam giác ABC có 3 góc nhọn , 2 đường cao BE và CF cắt nhau tại H

a/ Chứng minh tam giác AEB ~ tam giác AFC

b/ chứng minh tam giác DEF ~ tam giác ABC

c/ Tia AH cắt BC tại D. Chứng minh FC là tia phân giác góc DFE ?

\(A\)xác định \(\Leftrightarrow x^2y^2+1+\left(x^2-y\right)\left(1-y\right)\ne0\)

\(\Leftrightarrow x^2y^2+1+x^2-x^2y-y+y^2\ne0\)

\(\Leftrightarrow\left(x^2y^2+y^2\right)+\left(x^2+1\right)-\left(x^2y+y\right)\ne0\)

\(\Leftrightarrow y^2\left(x^2+1\right)+\left(x^2+1\right)-y\left(x^2+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left(y^2-y+1\right)\ne0\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\)

Ta có: \(\hept{\begin{cases}x^2+1>0\forall x\\\left(y-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall y\end{cases}}\)\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]>0\forall x;y\)

\(\Leftrightarrow\left(x^2+1\right)\left[\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ne0\forall x;y\)

\(\Leftrightarrow A\ne0\forall x;y\)

thiếu đề : \(\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}.\)

Bài 2 :

a, Để \(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4^2-4}{5}\)

\(\Rightarrow\hept{\begin{cases}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{cases}}\Rightarrow\orbr{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

b,\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)\frac{4x^2-4}{5}\)

\(B=\left[\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x+1\right)\left(x-1\right)}-\frac{x+3}{2\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{x^2+2x-3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\left[\frac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{4}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{8}{5}\)

=> giá trị của B ko phụ thuộc vào biến x

bài 1

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

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

=\(\left(4x\right)^2\)

=\(16x^2\)

Tại x=100 thay vào biểu thức trên ta có:

16*100^2=1600000

\(B=\left(\frac{x+1}{2x-2}+\frac{3}{x^2-1}-\frac{x+3}{2x+2}\right)=\left[\frac{x+1}{2.\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2.\left(x+1\right)}\right]\)

\(\Rightarrow\hept{\begin{cases}x\ne1\\x\ne\pm1\\x\ne-1\end{cases}\Rightarrow x\pm1}\)

Vậy để B xác định => x=+-1

ĐKXĐ: \(x\ne0;x\ne\pm2\)

a, \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(=\left[\frac{3x^2}{3x\left(x-2\right)\left(x+2\right)}-\frac{6x\left(x+2\right)}{3x\left(x-2\right)\left(x+2\right)}+\frac{3x\left(x-2\right)}{3x\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(=\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)

\(=\frac{-18x}{3x\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}\)

\(=\frac{-3x}{3x\left(x-2\right)}=\frac{-1}{x-2}\)

b, Ta có: \(\left|x\right|=\frac{1}{2}\Rightarrow x=\pm\frac{1}{2}\)

Với \(x=\frac{1}{2}\) thì \(A=\frac{-1}{\frac{1}{2}-2}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}\)

Với \(x=\frac{-1}{2}\)thì \(A=\frac{-1}{\frac{-1}{2}-2}=\frac{-1}{\frac{-5}{2}}=\frac{2}{5}\)

c, Để A=2 <=> \(\frac{-1}{x-2}=2\Leftrightarrow-1=2x-4\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Vậy x=3/2 thì A=2

d, Để A<0 <=> \(\frac{-1}{x-2}< 0\Leftrightarrow x-2>0\Leftrightarrow x>2\)

Vậy với x>2 thì A<0

e, Để A thuộc Z <=> x-2 thuộc Ư(-1)={1;-1}

Ta có: x-2=1 => x=3 (t/m)

          x-2=-1 => x=1 (t/m)

Vậy x thuộc {3;1} thì A thuộc Z

a)  \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)(ĐKXĐ: x khác 0; + 2)

\(A=\left(\frac{x^2}{x\left(x^2-4\right)}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)

\(A=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\right):\frac{6}{x+2}\)

\(A=\frac{-6x}{x\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-x}{x\left(x-2\right)}=\frac{1}{2-x}.\)

Vậy \(A=\frac{1}{2-x}.\)

b) \(\left|x\right|=\frac{1}{2}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\). Nếu \(x=\frac{1}{2}\)thì \(A=\frac{1}{2-\frac{1}{2}}=\frac{2}{3}.\)

Nếu \(x=-\frac{1}{2}\)thì \(A=\frac{1}{2+\frac{1}{2}}=\frac{2}{5}.\)Vậy ...

c) Để A=2 thì \(\frac{1}{2-x}=2\Rightarrow4-2x=1\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}.\)Vậy ...

d) Để A<0 thì \(\frac{1}{2-x}< 0\Rightarrow2-x< 0\Leftrightarrow x>2.\)Vậy ...

e) Để A thuộc Z thì \(\frac{1}{2-x}\in Z\Rightarrow1⋮2-x\). Mà 2-x thuộc Z (Do x thuộc Z)

Nên \(2-x\in\left\{1;-1\right\}\Rightarrow x\in\left\{1;3\right\}.\)(t/m ĐKXĐ)

Vậy x=1 hay x=3 thì A nguyên.

Ai giải giúp mình bài 1 với bài 4 trước đi

a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)

b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)

\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)

c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)

\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)

\(=\dfrac{3}{\sqrt{x}-2}\)