Question

Answer 1

a, đk x khác 3 ; -3 

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

\(=\dfrac{-x^2-6x-9+x^2-6x+9+4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x+3}\)

\(=\dfrac{4x^2-12x}{x-3}.\dfrac{1}{x+3}=\dfrac{4x}{x+3}\)

b, Ta có : \(2x^2-5x+2=0\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\Leftrightarrow x=2;x=\dfrac{1}{2}\)

Với x = 2 thì \(P=\dfrac{8}{5}\)

Với x = 1/2 thì \(\dfrac{2}{\dfrac{1}{2}+3}=\dfrac{2}{\dfrac{7}{2}}=2:\dfrac{7}{2}=\dfrac{4}{7}\)

Answer 2

Bn ơi tết rồi nghỉ đi bn ;-;