\(d.8x-24x^2+2-6x+18x^2-60x-3x+10-12=0\)
\(-6x^2-61x=0\)
\(-x\left(6x+61\right)=0\)
\(\left[{}\begin{matrix}-x=0\\6x+61=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-61}{6}\end{matrix}\right.\)
e: Ta có: \(\left(x-4\right)\left(x^2-4x+16\right)-\left(x^2+x\right)\left(x-4\right)=33x-66\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-4x+16-x^2-x\right)=33x-66\)
\(\Leftrightarrow\left(x-4\right)\left(-5x+16\right)=33x-66\)
\(\Leftrightarrow-5x^2+16x+20x-64-33x+66=0\)
\(\Leftrightarrow-5x^2+3x+2=0\)
\(\Leftrightarrow-5x^2+5x-2x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(-5x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
e) (x^2-4x+16)(x-4)-(x^2+x)(x-4)
=(x^2-4x+16-x^2+x)(x-4)
=(3x+16)(x-4)
=3x^2+4x-64
e. (x2 - 4x + 16)(x - 4) - (x2 + x)(x - 4) = 33x - 66
<=> (x2 - 4x + 16 - x2 - x)(x - 4) = 33x - 66
<=> (-5x + 16)(x - 4) = 33x - 66
<=> -5x2 + 20x + 16x - 64 = 33x - 66
<=> -64 + 66 = 5x2 - 20x - 16x
<=> 5x2 - 36x = 2
<=> x(5x - 36) = 2
<=> \(\left[{}\begin{matrix}x=1\\5x-36=2\\x=2\\5x-36=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7,6\\x=2\\x=7,4\end{matrix}\right.\)
