a: ĐKXĐ: \(y\notin\left\{0;2;-2;3\right\}\)
\(P=\left(\dfrac{2+y}{2-y}-\dfrac{4y^2}{y^2-4}-\dfrac{2-y}{2+y}\right):\dfrac{y^2-3y}{2y^2-y^3}:\dfrac{1}{y-3}\)
\(=\left(\dfrac{-\left(y+2\right)}{y-2}-\dfrac{4y^2}{\left(y-2\right)\left(y+2\right)}+\dfrac{y-2}{y+2}\right)\cdot\dfrac{y^2\left(2-y\right)}{y\left(y-3\right)}\cdot\dfrac{y-3}{1}\)
\(=\left(\dfrac{-\left(y+2\right)^2-4y^2+\left(y-2\right)^2}{\left(y-2\right)\left(y+2\right)}\right)\cdot\dfrac{-y\left(y-2\right)}{1}\)
\(=\dfrac{-y^2-4y-4-4y^2+y^2-4y+4}{\left(y-2\right)\left(y+2\right)}\cdot\dfrac{-y\left(y-2\right)}{1}\)
\(=\dfrac{-4y^2-8y}{\left(y+2\right)}\cdot\dfrac{-y}{1}=\dfrac{4y^2\left(y+2\right)}{y+2}=4y^2\)
b: \(2y^2-3y-2=0\)
=>\(2y^2-4y+y-2=0\)
=>(y-2)(2y+1)=0
=>\(\left[{}\begin{matrix}y-2=0\\2y+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=2\left(loại\right)\\y=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)
Khi y=-1/2 thì \(P=4\cdot\left(-\dfrac{1}{2}\right)^2=1\)
Lời giải:
a. ĐKXĐ: $y\neq 0; y\neq \pm 2; y\neq 3$
\(P=\left[\frac{-(y+2)^2}{(y-2)(y+2)}-\frac{4y^2}{(y-2)(y+2)}-\frac{-(y-2)^2}{(y+2)(y-2)}\right].\frac{2y^2-y^3}{y^2-3y}.(y-3)\)
\(=\frac{-(y+2)^2-4y^2+(y-2)^2}{(y-2)(y+2)}.\frac{y^2(2-y)(y-3)}{y(y-3)}\)
\(=\frac{-4y^2-8y}{(y-2)(y+2)}.\frac{-y^2(y-2)(y-3)}{y(y-3)}=\frac{-4y(y+2)}{(y-2)(y+2)}.-y(y-2)\)
\(=\frac{-4y}{y-2}.-y(y-2)=4y^2\)
b.
$2y^2-3y-2=0$
$\Leftrightarrow 2y(y-2)+(y-2)=0$
$\Leftrightarrow (y-2)(2y+1)=0$
$\Leftrightarrow y=2$ hoặc $y=\frac{-1}{2}$
Do $y\neq 2$ nên $y=\frac{-1}{2}$
Khi đó:
$P=4y^2=4(\frac{-1}{2})^2=1$
