a) (x+2)(x+3)-(x-2)(x+5)=0
\(x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(2x+16=0\)
\(2x=-16\)
\(x=-8\)
Vậy......
b) (8-5x)(x+2)+4(x-2)(x+1)+2(x-2)(x+2)=0
\(8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2+4x-4x-8=0\)
\(-6x+x^2=0\)
\(x\left(-6+x\right)=0\)
=> x=0 hoặc -6+x=0 <=>x=6
Vậy \(x\in\left\{0;6\right\}\)
a) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+2\right)x+\left(x+2\right).3-\left(x+5\right)x+\left(x+5\right).2=0\)
\(\Leftrightarrow x^2+2x+3x+6-x^2+5x+2x+10=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow12x=-16\)
\(\Leftrightarrow x=\frac{-4}{3}\)
Vậy...
