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3 tháng 4 2020

\(\frac{x+6}{x^2-4}-\frac{2}{x^2+2x}\)

\(=\frac{x+6}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x\left(x+2\right)}\)

\(=\frac{x\left(x+6\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)

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

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

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

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

\(\frac{x+6}{x^2+4}-\frac{2}{x^2+2x}\)

\(=\frac{x+6}{\left(x+2\right)^2}-\frac{2}{x\left(x+2\right)}\)

\(=\frac{x\left(x+6\right)}{x\left(x+2\right)^2}-\frac{2\left(x+2\right)}{x\left(x+2\right)^2}\)

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

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

20 tháng 12 2016

a. \(=\frac{x+1}{2.\left(x+3\right)}+\frac{2x+3}{x.\left(x+3\right)}=\frac{x^2+x+4x+6}{2x.\left(x+3\right)}=\frac{x^2+5x+6}{2x.\left(x+3\right)}=\frac{\left(x+2\right).\left(x+3\right)}{2x.\left(x+3\right)}=\frac{x+2}{2x}\)

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

Chắc chắn đúng, mik nhaaaaaa

6 tháng 11 2016

mk ko biết làm 

xin lỗi bn nhae

xin lỗi vì đã ko giúp được bn

chcus bn học gioi!

nhae@@@

6 tháng 11 2016

mình không biết làm

tk nhé@@@@@@@@@@@@@@@@@@@@

LOL

hihi

15 tháng 3 2020

1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)

=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)

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

15 tháng 3 2020

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

a) \(\frac{3x}{2x+4}+\frac{x+3}{x^2-4}\)

\(=\frac{3x}{2\left(x+2\right)}+\frac{x+3}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{3x\left(x-2\right)+2\left(x+3\right)}{2\left(x+2\right)\left(x-2\right)}\)

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

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

31 tháng 8 2020

\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)( ĐKXĐ : \(x\ne\pm2\))

\(=\frac{5\left(x+2\right)}{2\left(2x-4\right)}\cdot\frac{-\left(2x-4\right)}{x+2}\)

\(=\frac{-5\left(x+2\right)\left(2x-4\right)}{2\left(2x-4\right)\left(x+2\right)}\)

\(=-\frac{5}{2}\)

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}\)( ĐKXĐ : \(x\ne-5;x\ne6\))

\(=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\frac{3}{-\left(x-6\right)}\)

\(=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x+5\right)\left(x-6\right)}\)

\(=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}=-\frac{3x+18}{2x+10}\)

31 tháng 8 2020

a) 

Điều kiện : \(\hept{\begin{cases}4x-8\ne0\\x+2\ne0\end{cases}}\)    

\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)    

\(=\frac{5\left(x+2\right)}{-2\left(4-2x\right)}\cdot\frac{4-2x}{x+2}\)    

\(=\frac{-5}{2}\)    

b) 

Điều kiện : \(\hept{\begin{cases}2x+10\ne0\\6-x\ne0\end{cases}}\)    

\(\hept{\begin{cases}x\ne-5\\x\ne6\end{cases}}\)     

\(=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}\)   

\(=\frac{-6\left(x+6\right)\cdot3}{2x+10}\)   

\(=\frac{-9\left(x+6\right)}{x+5}\)  

\(=\frac{-9x-54}{x+5}\)  

\(=\frac{-9\left(x+5\right)-9}{x+5}\) 

\(=-9-\frac{9}{x+5}\)

2 tháng 8 2023

a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))

\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)

\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)

\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)

\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)

\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)

b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)

\(=1+\dfrac{3}{x}\)

\(=\dfrac{x}{x}+\dfrac{3}{x}\)

\(=\dfrac{x+3}{x}\)

a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)

\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)

b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)

3 tháng 9 2020

\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)( ĐKXĐ : \(x\ne1\))

\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)

\(=\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)

\(=\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x\left(x+5\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\frac{-x^2+5x-4}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\frac{-x^2+5x-4-\left(x^2+5x\right)}{x\left(x-1\right)\left(x^2+2\right)}\)

\(=\frac{-x^2+5x-4-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)

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

3 tháng 9 2020

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

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

Đang đánh máy thì bấm gửi -..-