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\(A=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(-2< x< 2\Leftrightarrow\left\{{}\begin{matrix}x-2< 0\\x+2>0\end{matrix}\right.\Leftrightarrow\left(x-2\right)\left(x+2\right)< 0;x\ne-1\Leftrightarrow\left(x+1\right)^2>0\Leftrightarrow A< 0\)
\(A=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x+1}{x^2-4}\)
a) \(A= \dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4} \\ =\dfrac{1}{x-2}+\dfrac{1}{x-2}+\dfrac{x^2+1}{(x-2)(x+2)} \\= \dfrac{x+2+x-2+x^2+1}{(x-2)(x+2)} \\=\dfrac{x^2+2x+1}{x^2-4} \\ =\dfrac{(x+1)^2}{(x-2)(x+2)}\)
b) Với mọi \(x\) thỏa mãn \(-2<x<2\) và \(x \ne -1\) thì \(x-2\) đều có giá trị âm, mà \(\begin{cases}(x+1)^2≥0\\x+2>0\\\end{cases}\) \( \Rightarrow\) Biểu thức A luôn có giá trị âm.
\(A=\frac{1}{x+2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Với \(\forall x\in\left[-2;2\right]\) thì \(\left(x-2\right)\left(x+2\right)< 0\Rightarrow\frac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}< 0\Rightarrow A< 0\)
\(a,A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\left(dkxd:x\ne\pm2\right)\)
\(=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2-4}\)
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
\(A=\dfrac{1}{x-2}+\dfrac{1}{x+2}+\dfrac{x^2+1}{x^2-4}\)
\(A=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{x+2+x-2+x^2+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\)
\(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
Ta có: -2 < x < 2
=> x thuộc { -1 ; 0 ; 1 }
Mà x khác -1 nên x = 0 ; x = 1
Với x = 0 thì \(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(0+1\right)^2}{\left(0-2\right)\left(0+2\right)}=\dfrac{1}{-4}\)
=> A có giá trị âm
Với x = 1 thì \(A=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(1+1\right)^2}{\left(1-2\right)\left(1+2\right)}=\dfrac{4}{-3}\)
=> A có giá trị âm
Vậy với -2 < x < 2 ; x khác -1 thì A có giá trị âm
a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)
b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)
c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)
\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)
P<=2
=>x+1>0
=>x>-1
Với `x \ne -5,x \ne -1` có:
`A=[x+2]/[x+5]+[-5x-1]/[x^2+6x+5]-1/[1+x]`
`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+5)(x+1)]`
`A=[x^2+x+2x+2-5x-1-x-5]/[(x+5)(x+1)]`
`A=[x^2-3x-4]/[(x+5)(x+1)]`
`A=[(x-4)(x+1)]/[(x+5)(x+1)]`
`A=[x-4]/[x+5]`
\(=\dfrac{x+2}{x+5}+\dfrac{-5x-1}{x^2+x+5x+5}-\dfrac{1}{x+1}\\ =\dfrac{x+2}{x+5}+\dfrac{-5x-1}{\left(x^2+x\right)+\left(5x+5\right)}-\dfrac{1}{x+1}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{x\left(x+1\right)+5\left(x+1\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}+\dfrac{-5x-1}{\left(x+1\right)\left(x+5\right)}-\dfrac{x+5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+2x+x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x^2+x-4x-4}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x^2+x\right)-\left(4x+4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x\left(x+1\right)-4\left(x+1\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+1\right)\left(x+5\right)}\\ =\dfrac{x-4}{x+5}\)
Lời giải:
1)
Ta có: \(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}+\frac{x^2+1}{x^2-4}\)
\(=\frac{x+2}{x^2-4}+\frac{x-2}{x^2-4}+\frac{x^2+1}{x^2-4}=\frac{x+2+x-2+x^2+1}{x^2-4}\)
\(=\frac{x^2+2x+1}{x^2-4}=\frac{(x+1)^2}{x^2-4}\)
2) Với mọi \(-2< x< 2\Rightarrow (x-2)(x+2)< 0\Leftrightarrow x^2-4< 0\)
Mà \((x+1)^2>0\forall x\neq 1; -2< x< 2\) nên \(\frac{(x+1)^2}{x^2-4}< 0\)
Tức là biểu thức A luôn nhận giá trị âm. Ta có đpcm.