Question

\(6x^2-41x+48=0\)
Này r lm sao mn giúp mình vs ạ:3

Answer 1

\(\Leftrightarrow6x^2-9x-32x+48=0\)

\(\Leftrightarrow3x\left(2x-3\right)-16\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{16}{3}\end{matrix}\right.\)

Answer 2

\(x\) = 1,5; \(x\) = \(\dfrac{16}{3}\)

Answer 3

\(6x^2-41x+48=0\\\Leftrightarrow\left(2x-3\right)\left(3x-16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{16}{3}\end{matrix}\right.\)