\(\frac{x}{2x-2}\)+\(\frac{x^2+1}{2-2x^2}\)
K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

21 tháng 10 2017

\(\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}=\frac{x}{2\left(x-1\right)}+\frac{x^2+1}{2\left(1-x^2\right)}\)

                                        \(=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x^2-1\right)}\)

                                         \(=\frac{x}{2\left(x-1\right)}-\frac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)

                                          \(=\frac{x\left(x+1\right)-\left(x^2+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

                                           \(=\frac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)

                                             \(=\frac{x-1}{2\left(x-1\right)\left(x+1\right)}\)

                                             \(=\frac{1}{2\left(x+1\right)}\)

15 tháng 12 2018

\(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\frac{-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

9 tháng 6 2017

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

11 tháng 12 2018

a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)

11 tháng 12 2018

Bạn có thể giúp mình 2 câu còn lại dc kh ạ 

3 tháng 8 2016

\(B=\left(\frac{2x}{x-3}-\frac{x-1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\left(ĐK:x\ne\pm3\right)\)

\(=\frac{2x\left(x+3\right)-\left(x-1\right)\left(x-3\right)-x^2-1}{x^2-9}:\frac{x+3-x+1}{x+3}\)

\(=\frac{2x^2+6x-x^2+3x+x-3-x^2-1}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}\)

\(=\frac{10x-4}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x+3}{4}=\frac{10x-4}{4\left(x-3\right)}\)

3 tháng 8 2016

\(B=\left(\frac{2x}{x-3}-\frac{x+1}{x+3}+\frac{x^2+1}{9-x^2}\right):\left(1-\frac{x-1}{x+3}\right)\)
\(=\left[\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x+1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+1}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+3-x+1}{x+3}\right)\)
\(=\left(\frac{2x^2+6x-x^2+3x-x+3-x^2-1}{\left(x+3\right)\left(x-3\right)}\right):\frac{4}{x+3}\)
\(=\frac{8x-1}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{4}\)\(=\frac{8x-1}{4\left(x-3\right)}\)