bài 1 :

tự làm

giúp mình mọi người ơi

a, ĐKXĐ: x\(\ne\) 1;-1;2

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

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

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

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

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

=\(\frac{x-2}{x-1}\)

c, Khi x= -1

→A= \(\frac{-1-2}{-1-1}\)

= -3

Vậy khi x= -1 thì A= -3

Câu d thì mình đang suy nghĩ nhé, mình sẽ quay lại trả lời sau ^^

a,ĐKXĐ:x#1; x#-1; x#2

b,Ta có:

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

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

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

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

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

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

c,Tại x=-1 ,theo ĐKXĐ x#-1 \(\Rightarrow\)A không có kết quả

d,Để A có giá trị nguyên \(\Rightarrow\frac{x-2}{x+1}\)có giá trị nguyên

\(\Leftrightarrow x-2⋮x+1\)

\(\Leftrightarrow x+1-3⋮x+1\)

Mà \(x+1⋮x+1\Rightarrow3⋮x+1\)

\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)

Mà theo ĐKXĐ x#2\(\Rightarrow x\in\left\{0;-2;-4\right\}\)

Vậy \(x\in\left\{0;-2;-4\right\}\)thì a là số nguyên

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

\(=\dfrac{x+4x+8+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x+2\right)}{3\left(x+1\right)}\)

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

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

Bài 2:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

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

\(=\dfrac{\left(x+2\right)\left(x+1\right)+6x-3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x}\cdot\dfrac{1}{2-4x}+\dfrac{x^2-3x-1}{3x}\)

\(=\dfrac{-8x^2+2}{3x}\cdot\dfrac{1}{-4x+2}+\dfrac{x^2-3x-1}{3x}\)

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

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

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

c: Khi x=1 thì \(D=\dfrac{1-1}{3}=0\)

Bài 1:

a: \(2x^2-8x=0\)

=>\(x^2-4x=0\)

=>x(x-4)=0

=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b: \(\left(x+2\right)^2-x\left(x-1\right)=10\)

=>\(x^2+4x+4-x^2+x=10\)

=>5x+4=10

=>5x=6

=>\(x=\dfrac{6}{5}\)

c: \(x^3-6x^2+9x=0\)

=>\(x\left(x^2-6x+9\right)=0\)

=>\(x\left(x-3\right)^2=0\)

=>\(\left[{}\begin{matrix}x=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)