Rút gọn biểu thức A=( \(\frac{1-x^3}{1-x}-x\)) : \(\frac{1-x^2}{1-x-x^2+x^3}\) với x khác 1 và -1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
6 tháng 9 2017
\(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x+x^2}{1-x}.\frac{\left(1-x\right)-x^2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x+x^2\right)-x\left(1-x\right)}{1-x}.\frac{\left(1-x\right)\left(1-x^2\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\frac{\left(1-x\right)\left(1+x^2\right)}{1-x}.\frac{\left(1-x\right)\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1+x\right)}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
\(=-x^3+x^2-x+1\)
6 tháng 9 2017
Ta có : \(A=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\)
\(=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{\left(1-x\right)}-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-\left(x^2-x^3\right)}\)
\(=\left(\left(1+x+x^2\right)-x\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)-x^2\left(x-1\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(=\left(1+x^2\right):\frac{\left(1-x\right)\left(1+x\right)}{\left(1-x\right)\left(1-x\right)\left(x+1\right)}\)
\(=\left(1+x^2\right):\frac{1}{1-x}\)
\(=\left(1+x^2\right)\left(1-x\right)\)
KN
12 tháng 10 2020
a) Với \(x\ne\pm1\)thì \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{2}{x^2-1}-\frac{x}{x-1}+\frac{1}{x+1}\right)=\left(\frac{x^2+2x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}\right):\left(\frac{2}{x^2-1}-\frac{x^2+x}{x^2-1}+\frac{x-1}{x^2-1}\right)=\frac{4x}{x^2-1}:\frac{1-x^2}{x^2-1}=\frac{-4x}{x^2-1}\)b) \(x=\sqrt{3+\sqrt{8}}=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
Khi đó \(A=\frac{-4\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)^2-1}=\frac{-4\left(\sqrt{2}+1\right)}{2\left(\sqrt{2}+1\right)}=-2\)
c) \(A=\sqrt{5}\Leftrightarrow\frac{-4x}{x^2-1}=\sqrt{5}\Leftrightarrow\sqrt{5}x^2+4x-\sqrt{5}=0\)
Dùng công thức nghiệm của phương trình bậc hai tìm được \(x=\frac{\sqrt{5}}{5}\)hoặc \(x=-\sqrt{5}\)
SN
2 tháng 8 2015
a)\(\left(\frac{1-x^3+1-x-x}{1-x}\right):\frac{-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}=\left(\frac{-x^3-2x+2}{1-x}\right)\cdot\left(1-x\right)=-x^3-2x+2\)
b) \(-\left(-1\frac{2}{3}\right)-2\cdot\left(-1\frac{2}{3}\right)+2=\frac{5}{3}+\frac{10}{3}+2=7\)
23 tháng 8 2020
Mình làm tắt thôi nhé
\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{\left(x+1\right)^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x^2-x+1\right)}=\frac{\left(x-1\right)^2}{x^2-x+1}\left(x\ne-1\right)\)
Dễ thấy \(A\ge0\)
KN
23 tháng 8 2020
\(A=\frac{x^4-2x^2+1}{x^4+x^3+x+1}=\frac{x^4-2x^3+x^2+2x^3-4x^2+2x+x^2-2x+1}{x^4-x^3+x^2+2x^2-2x^2+2x+x^2-x+1}\)
\(=\frac{x^2\left(x^2-2x+1\right)+2x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)}{x^2\left(x^2-x+1\right)+2x\left(x^2-x+1\right)+\left(x^2-x+1\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x^2-2x+1\right)}{\left(x^2+2x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^2-2x+1}{x^2-x+1}\)
\(=\frac{\left(x-1\right)^2}{x^2-x+1}\)
Ta có : \(\frac{\left(x-1\right)^2}{x^2-x+1}=\frac{\left(x-1\right)^2}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}\ge0\)
=> Đpcm
11 tháng 8 2016
Có: \(P=A:B=\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right):\left(1-\frac{2x}{x^2+1}\right)\left(ĐK:x\ne1\right)\)
\(=\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]:\left(\frac{x^2+1-2x}{x^2+1}\right)\)
\(=\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]:\frac{\left(x-1\right)^2}{x^2+1}\)
\(=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}\cdot\frac{x^2+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}\cdot\frac{x^2+1}{\left(x-1\right)^2}=\frac{1}{x-1}\)
b) Để \(P>-\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}>-\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{1}{x-1}+\frac{1}{2}>0\)
\(\Leftrightarrow\)\(\frac{2+x-1}{2\left(x-1\right)}>0\)
\(\Leftrightarrow\)\(\frac{x+1}{2\left(x-1\right)}>0\)
\(\Leftrightarrow\begin{cases}x+1>0\\2\left(x-1\right)>0\end{cases}\) hoặc \(\begin{cases}x+1< 0\\2\left(x-1\right)< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>-1\\x>1\end{cases}\) hoặc \(\begin{cases}x< -1\\x< 1\end{cases}\)
\(\Leftrightarrow x>1\) hoặc \(x< -1\)
\(A=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\frac{1-x^2}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(=\left(1+x+x^2-x\right):\frac{1}{1-x}=\left(1+x^2\right)\left(1-x\right)\)