a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

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

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)

a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)

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

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

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

\(\Leftrightarrow P=\frac{-1}{x+2}\)

b) Khi \(\left|x\right|=\frac{3}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)

c) Để P = 7

\(\Leftrightarrow-\frac{1}{x+2}=7\)

\(\Leftrightarrow7\left(x+2\right)=-1\)

\(\Leftrightarrow7x+14=-1\)

\(\Leftrightarrow7x=-15\)

\(\Leftrightarrow x=-\frac{15}{7}\)

Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)

d) Để \(P\inℤ\)

\(\Leftrightarrow1⋮x+2\)

\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{-3;-1\right\}\)

Vậy để  \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)

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a, \(E=\left(\frac{x^2+4}{x^2-4}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{x^2+4}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

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

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

b, Ta có : \(\left|2x-3\right|=1\Leftrightarrow\orbr{\begin{cases}2x-3=1\\2x-3=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(ktmđk\right)\\x=1\end{cases}}}\)

Thay x = 1 vào biểu thức E ta được : \(\frac{1+1}{6}=\frac{2}{6}=\frac{1}{3}\)

Vậy với x = 1 thì E = 1/3 

c, Ta có : \(E< 0\)hay \(\frac{x+1}{6}< 0\Rightarrow x+1>0\)( do 6 > 0 )

\(\Leftrightarrow x>-1\)

Với với x > -1 thì E < 0 

d, Ta có E = 3 - x hay \(\frac{x+1}{6}=3-x\Rightarrow x+1=18-6x\Leftrightarrow7x=17\Leftrightarrow x=\frac{17}{7}\)

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