Question

Tìm x, biết

a) 3x³ - 6x² = 0

b) x ( x-4) - 12x + 48 =0

c) x (x-4)- (x²-8) =0

d) 2x (x-5 ) - x (2x+3) =16

e) (4x²-1) - (x-1)² = -3

Answer 1

https://i.imgur.com/h3aQIDb.jpg

Answer 2

a. \(3x^3-6x^2=0\Leftrightarrow3x^2\left(x-2\right)=0\)

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

b. \(x\left(x-4\right)-12x+48=0\)

\(\Leftrightarrow x\left(x-4\right)-12\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=12\end{matrix}\right.\)

V.v.v.v