\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x...">
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9 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\frac{x-14}{x+3}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}=\frac{1}{x-3}\)

14 tháng 12 2018

\(\left(\frac{1}{x^2-9}+\frac{2}{3-x}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2}{x-3}+\frac{3}{x+3}\right):\frac{x-14}{x+3}\)

\(=\left(\frac{1}{\left(x+3\right)\left(x-3\right)}+\frac{-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right).\frac{x+3}{x-14}\)

\(=\left(\frac{1-2x-6+3x-9}{\left(x-3\right)\left(x+3\right)}\right).\frac{x+3}{x-14}=\frac{x-14}{\left(x+3\right)\left(x-3\right)}.\frac{x+3}{x-14}\)

\(=\frac{1}{x-3}\)

31 tháng 8 2020

Bài làm:

đk: \(x\ne-3;x\ne1\)

Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)

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

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

\(=-\frac{x^2-2x+1}{2x+6}\)

31 tháng 8 2020

\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)

\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)

3 tháng 3 2020

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)

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

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

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

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

\(=\frac{-3}{x-3}\)

11 tháng 12 2019

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

\(=\frac{-2}{x^2}\)

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

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

\(=x\left(x-3\right)\)

\(c)=\frac{x+3}{\left(x-1\right)\left(x+1\right)}-\frac{1}{x\left(x+1\right)}\)

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

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

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

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

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

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

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

\(x\left(x+1\right)+x\left(x-3\right)=4x\)

\(x^2+x+x^2-3x=4x\)

\(2x^2-2x=4x\)

\(2x^2-2x-4x=0\)

\(2x\left(x-3\right)=0\)

\(2x=0\Leftrightarrow x=0\)

hoặc 

\(x-3=0\Leftrightarrow x=3\)

22 tháng 4 2020

b) \(ĐKXĐ:x\ne\pm4\)

\(5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)

\(\Leftrightarrow5+\frac{96}{x^2-16}=\frac{2x-1}{x+4}+\frac{3x-1}{x-4}\)

\(\Leftrightarrow\frac{5\left(x^2-16\right)}{x^2-16}+\frac{96}{x^2-16}=\frac{\left(2x-1\right)\left(x-4\right)}{x^2-16}+\frac{\left(3x-1\right)\left(x+4\right)}{x^2-16}\)

\(\Rightarrow5\left(x^2-16\right)+96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)

\(\Leftrightarrow5x^2-80+96=2x^2-9x+4+3x^2+11x-4\)

\(\Leftrightarrow5x^2-2x^2-3x^2+9x-11x=4-4+80-96\)

\(\Leftrightarrow-2x=-16\)\(\Leftrightarrow x=8\)( thoả mãn ĐKXĐ )

Vậy tập nghiệm của phương trình là: \(S=\left\{8\right\}\)

1 tháng 12 2016

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

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

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

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

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

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

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

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

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

\(=\frac{1}{x}\)