Question

Answer 1

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` 

Answer 2

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