a, ĐK : \(x\ne0;1\)
\(A=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x\left(x-1\right)}\right)\)
\(=\dfrac{x^2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x^2}{x-1}\)
b, Thay x = 3 vào A ta được : \(\dfrac{9}{2}\)
c, \(A=4\Rightarrow\dfrac{x^2}{x-1}=4\Rightarrow x^2=4x-4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)
d, \(A< 2\Rightarrow\dfrac{x^2}{x-1}-2< 0\Leftrightarrow\dfrac{x^2-2x+1}{x-1}< 0\Rightarrow x-1< 0\Leftrightarrow x>1\)
a,\(\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x+1}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}.\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
b, Thay x=3 vào bt A ta có :
\(\dfrac{x^2}{x-1}=\dfrac{3^2}{3-1}=4,5\)
Vậy giá trị của biểu thức A bằng 4,5 khi x=3
c, Để A=4 \(\Leftrightarrow\dfrac{x^2}{x-1}=4\Leftrightarrow x^2=4x-4\)
\(\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x=2\)
a) Ta có: \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x}{x-1}\)
b) Thay x=3 vào A, ta được:
\(A=\dfrac{3}{3-1}=\dfrac{3}{2}\)
c) Để A=4 thì x=4x-4
\(\Leftrightarrow x-4x=-4\)
\(\Leftrightarrow-3x=-4\)
hay \(x=\dfrac{4}{3}\)
d) Để A<2 thì A-2<0
\(\Leftrightarrow\dfrac{x}{x-1}-2< 0\)
\(\Leftrightarrow\dfrac{x-2\left(x-1\right)}{x-1}< 0\)
\(\Leftrightarrow\dfrac{x-2x+2}{x-1}< 0\)
\(\Leftrightarrow\dfrac{-x+2}{x-1}< 0\)
\(\Leftrightarrow\dfrac{x-2}{x-1}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< 1\end{matrix}\right.\)
e) Để A nguyên thì \(x⋮x-1\)
\(\Leftrightarrow1⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1\right\}\)
\(\Leftrightarrow x\in\left\{2;0\right\}\)
Kết hợp ĐKXĐ, ta được: x=2
