a,\(9x^2-72x=0\)
\(9x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}9x=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
Vậy...
a
\(9x^2-72x=0\\ \Leftrightarrow9x\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
b)
\(\left(16-4x\right)\left(x+3\right)-\left(x+1\right)\left(4-4x\right)=40\\ \Leftrightarrow\left(16x+48-4x^2-12x\right)-\left(4x-4x^2+4-4x\right)=40\\ \Leftrightarrow4x+44=40\\ \Leftrightarrow x=-1\)
\(\left(16-4x\right)\left(x+3\right)-\left(x+1\right)\left(4-4x\right)=40\)
\(16x-4x^2+48-12x-4x+4x^2-4+4x=40\)
\(4x+44=40\)
\(\Rightarrow x=-1\)
Vậy x=-1
a) 9x^2-72x
=9x(x-8)
b) (16−4x)(x+3)−(x+1)(4−4x)=40
16x−4x^2+48−12x−4x+4x^2−4+4x=40
4x+44=40
⇒x=−1
Vậy x=-1