Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)ĐKXĐ: x\(\ne\)1;x\(\ne\)-1
B=\(\frac{1}{4x-4}\)
b)
B=\(\frac{1}{8016}\)
c)
x=\(\frac{4007}{4008}\)
a,
\(\Leftrightarrow A=\left(\frac{x+1}{\left(x+1\right)\left(x-1\right)}+\frac{x}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{2x+1}{\left(x+1\right)\left(x-1\right)}\cdot\frac{\left(x+1\right)^2}{2x+1}\)
\(\Leftrightarrow A=\frac{x+1}{x-1}\)
b, dùng máy tính kq là-3
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 \(x< -1\)
\(A=\left(\frac{x-2}{2x-2}+\frac{3}{2x-2}-\frac{x+3}{2x+2}\right):\left(-1-\frac{x-3}{x+1}\right)\)
\(=\left(\frac{x-2}{2\left(x-1\right)}+\frac{3}{2\left(x-1\right)}+\frac{-\left(x+3\right)}{2\left(x+1\right)}\right):\left(-\frac{1}{1}+\frac{-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{\left(x-2\right)\left(x+1\right)+3\left(x+1\right)-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{-1\left(x+1\right)-\left(x-3\right)}{x+1}\right)\)
\(=\left(\frac{x^2-x^2+x+3x-2x-6+3+3}{2\left(x-1\right)\left(x+1\right)}\right):\left(\frac{x-1-x+3}{x+1}\right)\)
=\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}:\frac{2}{x+1}\)
\(=\frac{2x}{2\left(x-1\right)\left(x+1\right)}.\frac{x+1}{2}\)
\(=\frac{x}{2\left(x-1\right)}\)
b,Thayx=2005
\(\Rightarrow A=\frac{2005}{4008}\)