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đây ak  

\(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x-\dfrac{1}{5}=0\\x^2+\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=\dfrac{1}{5}\\x^2=-\dfrac{1}{2}\left(VLý\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{2}{15}\)

1 - 2/7 = 7/7 - 2/7 = 5/7

15/42 - 2/7 = 15/42 - 12/47 = 3/47

6/5 x 3 = 18/5

7/3 x 2/5 = 14/15

`1-2/7=7/7-1/7=6/7`

`15/42-2/7=15/42-12/42=3/42=1/14`

`6/5xx3=(6xx3)/5=18/5`

`7/3xx2/5=(7xx2)/(3xx5)=14/15`

1 - 2/7 = 7/7 - 2/7 = 5/7

15/42 - 2/7 = 15/42 - 12/47 = 3/47

6/5 x 3 = 18/5

7/3 x 2/5 = 14/15

\(\Leftrightarrow3x+6+x^2-3x+2=9\)

\(\Leftrightarrow x^2+8=9\)

hay \(x\in\left\{1;-1\right\}\)

ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{3}{x-2}+\dfrac{x-1}{x+2}=\dfrac{9}{x^2-4}\\ \Leftrightarrow\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9=0\\ \Leftrightarrow3x+6+x^2-x-2x+2-9=0\\ \Leftrightarrow x^2-1=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

`9/[x^2-4]=[x-1]/[x+2]+3/[x-2]`      `ĐK: x \ne +-2`

`<=>9/[(x-2)(x+2)]=[(x-1)(x-2)+3(x+2)]/[(x-2)(x+2)]`

    `=>9=x^2-2x-x+2+3x+6`

`<=>x^2=1`

`<=>x=+-1` (t/m)

Vậy `x=+-1`

\(\dfrac{9}{x^2-4}=\dfrac{x-1}{x+2}+\dfrac{3}{x-2}\left(đkxđ:x\ne\pm2\right)\\ \Leftrightarrow\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ \Rightarrow9=x^2-3x+2+3x+6\\ \Leftrightarrow x^2=1\\ \Leftrightarrow x^2=\pm1\left(TM\right)\)

Vậy PT có tập nghiệm \(S=\left\{-1;1\right\}\)

\(\Leftrightarrow x^2-3x+2+3x+6=9\)

\(\Leftrightarrow x^2=1\)

=>x=1 hoặc x=-1

a, \(x=\left(\dfrac{3}{5}+\dfrac{2}{5}\right):\dfrac{4}{5}=\dfrac{5}{4}\)

b, \(x=\left(-\dfrac{1}{2}-\dfrac{2}{5}\right).\dfrac{3}{4}=-\dfrac{27}{40}\)

a)\(x=\left(\dfrac{3}{5}+\dfrac{2}{5}\right):\dfrac{4}{5}=\dfrac{5}{4}\)

b)\(x=\left(-\dfrac{1}{2}-\dfrac{2}{5}\right)\times\dfrac{3}{4}=-\dfrac{27}{40}\)

5/4

- 27/40