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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

3 tháng 3 2020

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

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

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2;-2\right\}\end{matrix}\right.\)

Ta có: \(\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

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

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

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}\)

\(=\dfrac{-1}{x-2}\)

17 tháng 4 2021

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)ĐK : \(x\ne-2;2\)

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

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

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

\(=\dfrac{x^3-4x+x^2-2x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{x^3+x^2-6x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{x^3+x^2-6x+24}{6\left(x-4\right)\left(x-2\right)}=\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}\)

17 tháng 4 2021

P/s : mình thấy đề này cứ sai sai ở đâu ý ! 

b, Ta có : \(\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x^2-3x+6\right)-12\left(x-4\right)\left(x-2\right)}{6\left(x-4\right)\left(x-2\right)}=0\)

\(\Rightarrow x^3-11x^2+66x-72=0\)

7 tháng 12 2020

bạn viết thế này khó nhìn quá

26 tháng 11 2021

nhìn hơi đau mắt nhá bạn hoa mắt quá