a) Ta có: \(E=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
b: Ta có: |x-3|=2
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Thay x=5 vào E, ta được:
\(E=\dfrac{25}{5-1}=\dfrac{25}{4}\)
tiếp ý c
\(E=\dfrac{1}{2}=>\dfrac{x^2}{x-1}=\dfrac{1}{2}=>2x^2=x-1< =>2x^2-x+1=0\)
\(< =>2\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}+\dfrac{15}{16}\right)=0\)
\(< =>\left(x-\dfrac{1}{4}\right)^2+\dfrac{15}{16}=0\left(vo-li\right)\)=>pt vô nghiệm
d,\(E>1< =>\dfrac{x^2}{x-1}>1< =>\dfrac{x^2}{x-1}-1>0\)
\(< =>\dfrac{x^2-x+1}{x-1}>0< =>\dfrac{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{x-1}>0\)
\(< =>x-1>0< =>x>1\)
f,\(E=\dfrac{x^2}{x-1}=4+\dfrac{x^2-4x+4}{x-1}=4+\dfrac{\left(x-2\right)^2}{x-1}\ge4\)
dấu"=" xảy ra<=>x=2(tm)
