Question

[nam doan]

mong làm hộ

Answer 1

[Quỳnh Anh]

Trả lời:

Bài 1: 

a, \(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)\(\left(ĐKXĐ:x\ne0;x\ne1\right)\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x\left(x-1\right)}\right]\)

\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\left[\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{x}{x\left(x-1\right)}+\frac{2-x^2}{x\left(x-1\right)}\right]\)

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

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

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

b, \(\left|2x-5\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=3\\2x-5=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=8\\2x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\left(tm\right)\\x=1\left(ktm\right)\end{cases}}}\)

Thay x = 4 vào A, ta có:

\(A=\frac{4^2}{4-1}=\frac{16}{3}\)

c, \(A=4\)

\(\Leftrightarrow\frac{x^2}{x-1}=4\)

\(\Leftrightarrow\frac{x^2}{x-1}=\frac{4\left(x-1\right)}{x-1}\)

\(\Rightarrow x^2=4x-4\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy x = 2 thì A = 4

d, \(A< 2\)

\(\Leftrightarrow\frac{x^2}{x-1}< 2\)

\(\Leftrightarrow\frac{x^2}{x-1}-2< 0\)

\(\Leftrightarrow\frac{x^2-2\left(x-1\right)}{x-1}< 0\)

\(\Leftrightarrow\frac{x^2-2x+2}{x-1}< 0\)

\(\Leftrightarrow x-1< 0\) ( vì \(x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\forall x\) )

\(\Leftrightarrow x< 1\)

Vậy x < 1 thì A < 2