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a: ĐKXĐ: x<>-1
b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)
\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)
c: P=2
=>x^2-2x=2x+2
=>x^2-4x-2=0
=>\(x=2\pm\sqrt{6}\)
\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}\\ A=\dfrac{2x^2+32}{2\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+16}{x^2-4}\\ c,A=-3\Leftrightarrow-3x^2+12=x^2+16\\ \Leftrightarrow4x^2=-4\Leftrightarrow x\in\varnothing\)
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)
\(=\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(\dfrac{x+1-x-3}{x+1}\right)\)
\(=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{-2}{x+1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{-2}\)
\(=\dfrac{-2x+2}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{-2\left(x-1\right)}{2\left(x-1\right)}\cdot\dfrac{-1}{2}\)
\(=\dfrac{1}{2}\)
Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)
a, điều kiện xác định: x2 - 4 ≠ 0
⇔ x2 ≠ 4
⇔x ≠ 2 và x ≠ -2
b, A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)
=\(\dfrac{x^2-x\left(x+2\right)+2\left(x-2\right)}{x^2-4}\)
= \(\dfrac{x^2-x^2-2x+2x-4}{x^2-4}\)
= \(\dfrac{x^2-4}{x^2-4}\)
= 1
c, x=1 ⇒ A= \(\dfrac{1^2}{1^2-4}-\dfrac{1}{1-2}+\dfrac{2}{1+2}\)
= \(\dfrac{4}{3}\)
a) Điều kiện xác định:
A\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.⇔\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) Rút gọn:
A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)[do MTC là (x-2)(x+2)].
A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{x^2-\left(x^2+2x\right)+2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{x^2-x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\)
A= \(\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)