Question

bài 1 ạ

Answer 1

Bài 1:

a) \(ĐKXĐ:x\ne0;x\ne1\)

\(P=\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right):\dfrac{x}{x-1}\)

\(=\left(\dfrac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\dfrac{x}{x\left(x-1\right)}+\dfrac{2-x^2}{x\left(x-1\right)}\right):\dfrac{x}{x-1}\)

\(=\left(\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\right):\dfrac{x}{x-1}\)

\(=\dfrac{x+1}{x\left(x-1\right)}.\dfrac{x-1}{x}\)

\(=\dfrac{x+1}{x^2}\)

b) \(\left|2x-1\right|=3\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Thay \(x=2\) vào P ta được:

\(P=\dfrac{x+1}{x^2}=\dfrac{2+1}{2^2}=\dfrac{3}{4}\)

Thay \(x=-1\) vào P ta được:

\(P=\dfrac{x+1}{x^2}=\dfrac{-1+1}{\left(-1\right)^2}=0\)

c) \(P=\dfrac{x+1}{x^2}=\dfrac{\dfrac{1}{4}x^2+x+1-\dfrac{1}{4}x^2}{x^2}=\dfrac{\left(\dfrac{1}{2}x+1\right)^2}{x^2}-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(P=-\dfrac{1}{4}\Leftrightarrow x=-2\)

Vậy \(P_{min}=-\dfrac{1}{4}\)

 

Answer 2

lx

Answer 3