Question

[Việt Phương]

cho biểu thức M=(\(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\)) : (\(x-2+\frac{10-^{x^2}}{x+2}\))

a) rút gọn M 

b) tính giá trị của M khi /x/= \(\frac{1}{2}\)

Answer 1

[Nguyễn Xuân Anh]

a)\(\text{ĐKXĐ:}\hept{\begin{cases}x^3-4x\ne0\\6-3x\ne0\\x+2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne\mp2\end{cases}}\)

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

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

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

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

    \(=\frac{1}{x+2}\)

b) /x/= \(\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

*\(\text{Với }x=\frac{1}{2}\text{ta có pt:}\)

  \(M=\frac{1}{x+2}=\frac{1}{\frac{1}{2}+2}=\frac{2}{5}\)

*\(\text{Với x= -1/2 ta có pt:}\)

 \(M=\frac{1}{x+2}=\frac{1}{-\frac{1}{2}+2}=\frac{2}{3}\)

Answer 2

[phạm thị hồng nhung]

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

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

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

            =(\(\frac{3x^2-6x^2-12x+3x^2-6x}{3x\left(x-2\right)\left(x+2\right)}\)):(\(\frac{x^2-4+10-x^2}{x+2}\))

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

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

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

               =\(\frac{-1}{x-2}\)

  Vậy M=\(\frac{-1}{x-2}\)

b)Vì /x/ =1/2 nên x=1/2 hoặc x=-1/2Thay x=1/2 vào M ta được;

     \(\frac{-1}{\frac{1}{2}-2}\)=\(\frac{2}{3}\)

  Thay x=-1/2 vào M ta được:

\(\frac{-1}{-\frac{1}{2}-2}\)=\(\frac{2}{5}\)

    Vậy \(M\in\)\(\hept{\begin{cases}\\\end{cases}\frac{2}{5};\frac{2}{3}}\)khi /x/=1/2