a: \(\left(x^2+2x+4\right)\left(2-x\right)+x\left(x-3\right)\left(x+4\right)=x^2+24\)
=>\(2^3-x^3+x\left(x^2+x-12\right)-x^2-24=0\)
=>\(-x^3-x^2-16+x^3+x^2-12x=0\)
=>-12x-16=0
=>12x=-16
=>\(x=-\dfrac{4}{3}\)
b: \(\left(\dfrac{x}{2}+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\dfrac{x}{4}+3\right)=0\)
=>\(\dfrac{5}{2}x-3x^2+15-18x+3x^2+36x-\dfrac{1}{2}x-6=0\)
=>20x+9=0
=>\(x=-\dfrac{9}{20}\)
c: \(\left(x-3\right)^2+\left(x+3\right)^2-x\left(x+5\right)=18\)
=>\(x^2-6x+9+x^2+6x+9-x^2-5x=18\)
=>\(x^2-5x+18=18\)
=>\(x^2-5x=0\)
=>x(x-5)=0
=>\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a, \(-\left(x-2\right)\left(x^2+2x+4\right)+x\left(x-3\right)\left(x+4\right)=x^2+24\)
\(\Leftrightarrow-\left(x^3-8\right)+x\left(x^2+x-12\right)=x^2+24\)
\(\Leftrightarrow-x^3+8+x^3+x^2-12x=x^2+24\)
\(\Leftrightarrow-12x=16\Leftrightarrow x=-\dfrac{4}{3}\)
b, \(\left(\dfrac{x}{2}+3\right)\left(5-6x\right)+\left(12x-2\right)\left(\dfrac{x}{4}+3\right)=0\)
\(\Leftrightarrow\dfrac{5x}{2}-3x^2+16-18x+3x^2+36x-\dfrac{x}{2}-6=0\)
\(\Leftrightarrow2x+18x+10=0\Leftrightarrow x=-\dfrac{1}{2}\)
c, \(\left(x-3\right)^2+\left(x+3\right)^2-x\left(x+5\right)=18\)
\(\Leftrightarrow x^2-6x+9+x^2+6x+9-x^2-5x=18\Leftrightarrow x^2-5x=0\Leftrightarrow x=0;x=5\)
