Question

Bài 1 : Tính

a) (2x + 1)(4x^2 - 6x + 9)

b) (x - 2/3y)(x^2 + 2/3xy + 4/9y^2)

giải rõ giúp mình vs ạ!

Answer 1

a,Đặt \(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)^2=0\Leftrightarrow x=-\dfrac{1}{2};x=\dfrac{3}{2}\)

b,sửa đề \(\Leftrightarrow\left(x-\dfrac{2}{3}y\right)\left(x^2+\dfrac{4}{3}xy+\dfrac{4}{9}y^2\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{3}y\right)\left(x+\dfrac{2}{3}y\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}y\\x=-\dfrac{2}{3}y\end{matrix}\right.\)