Cho A=\(\left(\frac{x}{x-2}+\frac{2x}{x^2-2x}\right)\left(x^2+4\right)\)
Rút gọn
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
31 tháng 10 2020
Bài làm
Như đã nhắn là mình sẽ làm theo quan điểm của mình là 5/(x^2 - 1) nha
\(A=\left[\frac{3\left(x+2\right)}{2x^3+2x+2x^2+2}+\frac{2x^2-x-10}{2x^3-2-2x^2+2x}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2x+2}-\frac{3}{2x-2}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{2x^2\left(x+1\right)+2\left(x+1\right)}+\frac{2x^2+4x-5x-10}{\left(2x^3-2x^2\right)+\left(2x-2\right)}\right]:\left[\frac{5}{x^2-1}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{2x\left(x+2\right)-5\left(x+2\right)}{2x^2\left(x-1\right)+2\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)}{\left(2x^2+2\right)\left(x+1\right)}+\frac{\left(2x-5\right)\left(x+2\right)}{\left(2x^2+2\right)\left(x-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3}{2\left(x+1\right)}-\frac{3}{2\left(x-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}+\frac{\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{5\cdot2}{2\left(x+1\right)\left(x-1\right)}+\frac{3\left(x-1\right)}{2\left(x^2-1\right)}-\frac{3\left(x+1\right)}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{3\left(x+2\right)\left(x-1\right)+\left(2x-5\right)\left(x+2\right)\left(x+1\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10}{2\left(x^2-1\right)}+\frac{3x-3}{2\left(x^2-1\right)}-\frac{3x+3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left[3x-3+\left(2x-5\right)\left(x+1\right)\right]}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\left[\frac{10+3x-3-3x-3}{2\left(x^2-1\right)}\right]\)
\(A=\left[\frac{\left(x+2\right)\left(3x-3+2x^2+2x-5x-5\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\right]:\frac{4}{2\left(x^2-1\right)}\)
\(A=\frac{\left(x+2\right)\left(2x^2-8\right)}{\left(2x^2+2\right)\left(x^2-1\right)}\cdot\frac{\left(x^2-1\right)}{2}\)
\(A=\frac{\left(x+2\right)2\left(x^2-4\right)}{2\left(2x^2+2\right)}\)
\(A=\frac{2\left(x+2\right)\left(x-2\right)\left(x+2\right)}{4\left(x^2+1\right)}\)
\(A=\frac{\left(x+2\right)^2\left(x-2\right)}{2\left(x^2+1\right)}\)
:>>> Chả biết đúng không nữa nhưng số to quá :>>
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
19 tháng 11 2016
\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{\left(x^2+xy+y^2\right)}\)
\(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{\left(x-2\right)3\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
\(\frac{2x^3+x^2-2x-1}{x^3+2x^2-x-2}=\frac{\left(x-1\right)\left(x+1\right)\left(2x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=\frac{2x+1}{x+2}\)
19 tháng 11 2016
\(\frac{x^4-y^4}{y^3-x^3}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(y-x\right)\left(x^2+xy+y^2\right)}=-\frac{\left(x^2+y^2\right)\left(x+y\right)}{\left(x^2+xy+y^2\right)}\)
25 tháng 12 2020
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
\(ĐKXĐ:\hept{\begin{cases}x-2\ne0\\x^2-2x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\left(x-2\right)\ne0\Leftrightarrow x\ne0\end{cases}.}}\)
\(A=\left(\frac{x}{x-2}+\frac{2x}{x^2-2x}\right)\left(x^2+4\right)\)
\(=\left(\frac{x}{x-2}+\frac{2x}{x\left(x-2\right)}\right)\left(x^2+4\right)\)
\(=\frac{x+2}{x-2}.\left(x^2+4\right)\)(x^2-4 còn rút gọn đc thế này thì bó tay)
( Sai dấu )
ĐKXĐ
\(\hept{\begin{cases}x-2\ne0\\x\left(x-2\right)\ne0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x\ne2\\x\ne0\end{cases}}\) ( T/m đk )
\(A=\left(\frac{x}{x-2}+\frac{2x}{x^2-2x}\right).\left(x^2-4\right)\)
\(=\left[\frac{x}{x-2}+\frac{2x}{x\left(x-2\right)}\right]\left(x^2-4\right)\)
\(=\left[\frac{x}{x-2}+\frac{2}{\left(x-2\right)}\right]\left(x^2-4\right)\)
\(=\frac{x+2}{x-2}.\left(x^2-4\right)\)
\(=\frac{x+2.x^2+4}{x+2}=\frac{x+2}{x-2}.\left(x-2\right)\left(x+2\right)\)
\(=\frac{x+2\left(x-2\right)\left(x+2\right)}{\left(x-2\right)}=\left(x+2\right)^2\)