Question

  Hộ e bài 6 vs a

Answer 1

6: ĐKXĐ: y<>0; y<>2; y<>-2; y<>3

a: \(P=\left(\dfrac{-\left(y+2\right)}{y-2}-\dfrac{4y^2}{\left(y-2\right)\left(y+2\right)}+\dfrac{y-2}{y+2}\right):\dfrac{y\left(y-3\right)}{y^2\left(2-y\right)}:\dfrac{1}{y-3}\)

\(=\dfrac{-y^2-4y-4-4y^2+y^2-4y+4}{\left(y-2\right)\left(y+2\right)}\cdot\dfrac{y\left(2-y\right)}{y-3}\cdot\dfrac{y-3}{1}\)

\(=\dfrac{-4y^2-8y}{\left(y-2\right)\left(y+2\right)}\cdot\dfrac{-y\left(y-2\right)}{1}\)

\(=4y^2\)

b: 2y^2-3y-2=0

=>2y^2-4y+y-2=0

=>(y-2)*(2y+1)=0

=>y=2(loại) hoặc y=-1/2(nhận)

Khi y=-1/2 thì P=4*(-1/2)^2=1