a ) \(D=\left(\dfrac{1}{1-x}+\dfrac{1}{1+x}\right):\left(\dfrac{1}{1-x}-\dfrac{1}{1+x}\right)+\dfrac{1}{x+1}\)
\(=\left(\dfrac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}\right):\left(\dfrac{1+x-1+x}{\left(1-x\right)\left(1+x\right)}\right)+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}:\dfrac{2x}{\left(1-x\right)\left(1+x\right)}+\dfrac{1}{x+1}\)
\(=\dfrac{2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(1-x\right)\left(1+x\right)}{2x}+\dfrac{1}{x+1}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}\)
\(=\dfrac{x+1+x}{x\left(x+1\right)}=\dfrac{2x+1}{x\left(x+1\right)}\)
b ) Khi \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Thay 0,1 vào biểu thức D
Khi \(x=0\), ta có :
\(\dfrac{2.0+1}{0.\left(0+1\right)}\) ( ko được )
Khi \(x=1,\) ta có :
\(\dfrac{2.1+1}{1.\left(1+1\right)}=\dfrac{3}{2}\)
c ) Khi \(D=\dfrac{3}{2}\)
Ta có : \(\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{3}{2}\)
\(\Leftrightarrow4x+2=3x^2+3x\)
\(\Leftrightarrow-3x^2+x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ...........
