Bài 10;
\(a,ĐK:x\ne1;x\ne0\\ P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\\ =\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right]\\ =\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\\ =\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x+1}{x\left(x-1\right)}\\ =\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\\ =\dfrac{x^2}{x-1}\)
b) \(P< 1=>\dfrac{x^2}{x-1}< 1=>\dfrac{x^2}{x-1}-1< 0\)
\(=>\dfrac{x^2-x+1}{x-1}< 0\)
Mà: `x^2-x+1=(x^2-2*x*1/2+1/4)+3/4=(x-1/2)^2+3/4>=3/4>0`
\(=>x-1< 0=>x< 1\)
Kết hợp với đkxđ: `x<1` và `x<>0`
\(c,x>1=>x-1>0\\ =>P=\dfrac{x^2}{x-1}\ge0\)
Vì `x^2>=0` với mọi x
Dấu "=" xảy ra: `x^2=0<=>x=0`
Bài 3:
a: \(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(=\left(-\dfrac{1}{x-1}+\dfrac{2}{x+1}-\dfrac{x-5}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x^2-1}{-2x+1}\)
\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{-2x+1}\)
\(=\dfrac{4}{-2x+1}\)
Bài 9:
a ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
\(P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x^2-1+x+2-x^2}\)
\(=\dfrac{x^2\left(x+1\right)}{x-1}\cdot\dfrac{1}{x+1}=\dfrac{x^2}{x-1}\)
b: Để \(P=-\dfrac{1}{2}\) thì \(\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)
=>\(2x^2=-x+1\)
=>\(2x^2+x-1=0\)
=>\(2x^2+2x-x-1=0\)
=>(x+1)(2x-1)=0
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)
c: Để P nguyên thì \(x^2⋮x-1\)
=>\(x^2-1+1⋮x-1\)
=>\(1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>\(x\in\left\{2;0\right\}\)
Kết hợp ĐKXĐ, ta được: x=2
d: \(P=\dfrac{x^2}{x-1}=\dfrac{x^2-1}{x-1}+\dfrac{1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\)
=>\(P>=2\cdot\sqrt{\left(x-1\right)\cdot\dfrac{1}{x-1}}+2=4\)
Dấu '=' xảy ra khi x-1=1
=>x=2
