\(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}\) `ĐKXĐ : x ≠ \(\pm1\)`
1)\(\Leftrightarrow A=\left(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{5x-5}{2x}\)
\(\Leftrightarrow A=\left(\dfrac{x^2+2x+1-x^2+2x-1}{x^2-1}\right)\cdot\dfrac{5x-5}{2x}\)
\(\Leftrightarrow A=\dfrac{20x^2-20x}{2x^3-2x}=\dfrac{2x\left(10x-10\right)}{2x\left(x^2-1\right)}=\dfrac{10\left(x-1\right)}{x^2-1}\)
3) thay x = -3 vào A ta đc
\(A=\dfrac{10.\left(-3-1\right)}{\left(-3\right)^2-1}=\dfrac{-40}{9-1}=-\dfrac{40}{8}=-5\)
4)\(\left|x-2\right|=4-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=4-2x\\x-2=-4+2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2-4+2x=0\\x-2+4-2x=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x-6=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
\(=>A=\dfrac{10\left(2-1\right)}{2^2-1}=\dfrac{10}{4-1}=\dfrac{10}{3}\)
Sai thì thoi nha ( e lớp 7 nên ko chắc)
5)\(A=\dfrac{10x-10}{x^2-1}=2\)
\(\Leftrightarrow10x-10-2x^2+2=0\)
`<=> -2x^2+10x-8=0`
`<=> -2(x^2 -5x+4)=0`
`<=> -2(x-4)(x-1)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
6)\(\dfrac{10x-10}{x^2-1}< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}10x-10< 0\\x^2-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10x< 10\\x^2< 1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x< \pm1\end{matrix}\right.\)
7.
\(A=\dfrac{10}{x+1}\)
Để `A` nguyên thì \(x+1\in U\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
`@x+1=1->x=0(tm)`
`@x+1=-1->x=-2(tm)`
`@x+1=2->x=1(ktm)`
`@x+1=-2->x=-3(tm)`
`@x+1=5->x=4(tm)`
`@x+1=-5->x=-6(tm)`
`@x+1=10->x=9(tm)`
`@x+1=-10->x=-11(tm)`
Vậy \(x\in\left\{0;-2;-3;4;-6;9;-11\right\}\) thì `A` đạt giá trị nguyên
8.
\(A>-1\)
\(\Leftrightarrow\dfrac{10}{x+1}>-1\)
\(\Leftrightarrow\dfrac{10}{x+1}+1>0\)
\(\Leftrightarrow\dfrac{10+x+1}{x+1}>0\)
\(\Leftrightarrow\dfrac{x+11}{x+1}>0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+11>0\\x+1>0\end{matrix}\right.\) hoặc \(\Leftrightarrow\left\{{}\begin{matrix}x+11< 0\\x+1< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-11\\x>-1\end{matrix}\right.\) | \(\Leftrightarrow\left\{{}\begin{matrix}x< -11\\x< -1\end{matrix}\right.\)
Vậy \(x>-1\) hoặc \(x< -11\) thì `A<-1`
