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\(a,A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2-4}\)
Vậy \(A=\dfrac{\left(x+1\right)^2}{x^2-4}\)
\(b,\) Theo đề, ta có : \(-2< x< 2\)
\(\Rightarrow x-2< 0;x+2>0;\left(x+1\right)^2>0\)
\(\Rightarrow A< 0\) hay phân thức luôn có giá trị âm
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{-4x}{\left(x+4\right)\left(x-1\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Sửa đề: 4/x^2-1
a: \(A=\left(\dfrac{x+1}{x-1}+\dfrac{4}{x^2-1}-\dfrac{x-1}{x+1}\right):\dfrac{x^2-4x+4}{x^2+x}\)
\(=\dfrac{x^2+2x+1+4-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x+1\right)}{\left(x-2\right)^2}\)
\(=\dfrac{4x+4}{\left(x-1\right)}\cdot\dfrac{x}{\left(x-2\right)^2}=\dfrac{x\left(4x+4\right)}{\left(x-1\right)\left(x-2\right)^2}\)
b: Khi x=1/2 thì \(A=\dfrac{\dfrac{1}{2}\left(2+4\right)}{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}-2\right)^2}=\dfrac{-8}{3}\)
a: \(A=\left(\dfrac{4}{x}-1\right):\left(1-\dfrac{x-3}{x^2+x+1}\right)\)
\(=\dfrac{4-x}{x}:\dfrac{x^2+x+1-x+3}{x^2+x+1}\)
\(=\dfrac{4-x}{x}\cdot\dfrac{x^2+x+1}{x^2+4}=\dfrac{\left(4-x\right)\left(x^2+x+1\right)}{x\left(x^2+4\right)}\)
b: x^4-7x^2-4x+20=0
=>(x-2)^2(x^2+4x+5)=0
=>x=2
Khi x=2 thì \(A=\dfrac{\left(4-2\right)\left(4+2+1\right)}{2\left(4+4\right)}=\dfrac{7}{8}\)