Rút gọn A = \(\frac{x^2}{x^2-1}-\frac{x^2}{x^2+1}\left(\frac{x}{x+1}+\frac{1}{x^2+x}\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\left(\dfrac{x^2-2x+1}{x^2+x+1}-\dfrac{-2x^2+4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(=\dfrac{x^3-3x^2+3x-1+2x^2-4x-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}=\dfrac{x^2+1}{2}\)
\(A=\left(\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right)\left(ĐK:x\ge0;\ne1\right)\)
\(=\left[\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\left[\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right]\)
\(=\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
\(=\frac{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=\frac{2\left(\sqrt{x}+3\right)}{\sqrt{x}+1}\)
\(=\dfrac{a+x+1}{a+x}:\dfrac{a+x-1}{a+x}\cdot\left(\dfrac{2ax-1+a^2+x^2}{2ax}\right)\)
\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x\right)^2-1}{2ax}\)
\(=\dfrac{a+x+1}{a+x-1}\cdot\dfrac{\left(a+x+1\right)\left(a+x-1\right)}{2ax}\)
\(=\dfrac{\left(a+x+1\right)^2}{2ax}\)
\(=\dfrac{x^3-1}{x}\cdot\dfrac{x^2-1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x^2+x}{x}=2x+1\)
\(A=\left(\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}+\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}\right):\dfrac{x-1}{x^3}\)
\(=\dfrac{x^2+3}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)^2}\)
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
ĐKXĐ:\(\sqrt{x}\ge0\Leftrightarrow x\ge0\)
Rút gọn: P=\(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\frac{\left(1-x\right)^2}{2}=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(x-1\right)^2}{2}=\frac{2\sqrt{x}\left(x-1\right)^2}{2\left(x-1\right)\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)=x-1\)
\(A=\frac{x^2}{x^2-1}-\frac{x^2}{x^2+1}\left(\frac{x}{x+1}+\frac{1}{x^2+x}\right)\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x}{x+1}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}\left[\frac{x^2}{x\left(x+1\right)}+\frac{1}{x\left(x+1\right)}\right]\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x^2}{x^2+1}.\frac{x^2+1}{x\left(x+1\right)}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x}{x+1}\)
=>\(A=\frac{x^2}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
=>\(A=\frac{x^2-x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x^2-x^2+x}{\left(x-1\right)\left(x+1\right)}\)
=>\(A=\frac{x}{x^2-1}\)