Đáp án:
`D=x/(x^2+2x)+(2x)/(x^2-4)+1/(2-x)`
`ĐKXĐ:`
$\left\{\begin{matrix}x^2+2x\ne0& \\x^2-4\ne0&\\ 2-x\ne0& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x(x+2)\ne0& \\(x-2)(x+2)\ne0&\\ x\ne2& \end{matrix}\right.$
`=>` $\left\{\begin{matrix}x\ne0& \\x\ne-2&\\ x\ne2& \end{matrix}\right.$
`a) D=x/(x^2+2x)+(2x)/(x^2-4)+1/(2-x)`
`D=x/(x(x+2))+(2x)/((x-2)(x+2))-1/(x-2)`
`MTC: x(x+2)(x-2)`
`=(x.(x-2))/(x(x+2)(x-2))+(2x.x)/(x(x+2)(x-2))-(1.x(x+2))/(x(x+2)(x-2))`
`=(x^2-2x)/(x(x+2)(x-2))+(2x^2)/(x(x+2)(x-2))-(x^2+2x)/(x(x+2)(x-2))`
`=(x^2-2x+2x^2-x^2-2x)/(x(x+2)(x-2))`
`=(2x^2-4x)/(x(x+2)(x-2))`
`=(2x(x-2))/(x(x+2)(x-2))`
`=2/(x+2)`
`c) D=1/3 <=> 2/(x+2)=1/3`
`=> x+2=2.3=6`
`=> x=6-2`
`=> x=4 (tmđk)`
