\(1,ĐKXĐ:x\ne\pm1;x\ne0\)
\(2,A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}\)
\(=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\dfrac{5\left(x-1\right)}{2x}\)
\(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}.\dfrac{5\left(x-1\right)}{2x}\)
\(=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}.\dfrac{5\left(x-1\right)}{2x}\)
\(=\dfrac{10}{x+1}\)
3, Thay x = -3 vào A, ta có :
\(A=\dfrac{10}{x+1}=\dfrac{10}{-3+1}=\dfrac{10}{-2}=-5\)
4, Ta có : \(\left|x-2\right|=4-2x\)
* TH1 : nếu \(x-2\ge0\Leftrightarrow x\ge2\)
\(\Rightarrow x-2=4-2x\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\left(nhận\right)\)
* TH2 : nếu \(x-2< 0\Leftrightarrow x< 2\)
\(\Rightarrow-x+2=4-2x\)
\(\Leftrightarrow x=2\left(loại\right)\)
Thay x = 2 vào A , ta có :
\(A=\dfrac{10}{x+1}=\dfrac{10}{2+1}=\dfrac{10}{3}\)
5, Để A = 2 thì
\(A=\dfrac{10}{x+1}=2\)
\(\Leftrightarrow10=2x+2\)
\(\Leftrightarrow-2x=-8\)
\(\Leftrightarrow x=4\)
Vậy x = 4 thì A = 2
6, Để A < 0 thì \(\dfrac{10}{x+1}< 0\)
Mà 10 > 0
\(\Rightarrow x+1< 0\Leftrightarrow x< -1\)
7, Ta có : \(A=\dfrac{10}{x+1}\)
Để A nhận giá trị nguyên thì \(10⋮x+1\Rightarrow x+1\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
\(\Rightarrow x\in\left\{0;-2;1;-3;4;-6;9;-11\right\}\)
Mà \(x\ne\pm1;x\ne0\left(đkxđ;x\in Z\right)\)
\(\Rightarrow x\in\left\{-2;-3;4;-6;9;-11\right\}\)
8, Để A > -1 thì \(\dfrac{10}{x+1}>-1\)
\(\Leftrightarrow\dfrac{10}{x+1}+1>0\)
\(\Leftrightarrow\dfrac{10+x+1}{x+1}>0\)
\(\Leftrightarrow\dfrac{x+11}{x+1}>0\)
* TH1 : x + 11 > 0 ; x + 1 >0
=> x > -11 ; x > - 1 => x > - 1
* TH2 : x + 11 < 0 ; x + 1 < 0
=> x < -11 ; x < - 1 => x < -11
Vậy \(x< -11\) và \(x>-1\) thì \(A>-1\)
