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a: \(A=\dfrac{1}{x-1}-\dfrac{x^2+x}{x^2+1}\cdot\dfrac{x+1-x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
b: A=1/5
=>\(\dfrac{x-1}{x^2+1}=\dfrac{1}{5}\)
=>x^2+1=5x-5
=>x^2-5x+6=0
=>x=2 hoặc x=3
\(a,ĐK:x\ne1\\ b,A=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\\ c,A=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\left(tm\right)\)
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)
\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)
\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)
b: Thay x=-4 vào A, ta được:
\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)
\(a,ĐKXĐ:x\ne\pm1;x\ne-\frac{1}{2}\)
\(b,A=\left(\frac{1}{x+1}-\frac{2}{x-1}-\frac{x+5}{1-x^2}\right):\frac{2x+1}{x^2-1}\)
\(A=\left[\frac{x-1}{\left(x+1\right)\left(x-1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x+5}{\left(x+1\right)\left(x-1\right)}\right]:\frac{2x+1}{\left(x+1\right)\left(x-1\right)}\)
\(A=\left[\frac{x-1-2x-2+x+5}{\left(x+1\right)\left(x-1\right)}\right]:\frac{2x+1}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{2}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{2x+1}\)
\(A=\frac{2}{2x+1}\)
\(c,Để:A>0\)
\(\Rightarrow2x+1>0\)
\(\Rightarrow x>-\frac{1}{2}\)
\(Để:A< 0\)
\(\Rightarrow2x+1< 0\)
\(\Rightarrow x< -\frac{1}{2}\)
Vậy \(x>-\frac{1}{2}\) và \(x\ne1\) thì A>0
\(x< -\frac{1}{2}\) và \(x\ne-1\) thì A<0
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
hé lo