Question

giúp mình nhanh nha

Answer 1

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

\(b,\Leftrightarrow\left(x+2\right)\left(x-1\right)+3\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(c,\Leftrightarrow\left(3x-1\right)^2=\left(4x+6\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=4x+6\\3x-1=-4x-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{5}{7}\end{matrix}\right.\)