`-10x(2-x)-5x(x+2)=5x(x+3)`
`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`
`<=> 5x^2 - 30x = 5x^2 + 15x`
`<=> -30x = 15x`
`<=> -45x = 0`
`<=> x = 0`
Vậy `S = {0}`
\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)
\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)
\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)
\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)
\(\text{⇔}15x^2-30=-5x^2+15\)
\(\text{⇔}15x^2+5x^2=30+15\)
\(\text{⇔}20x^2=45\)
\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
Ta có: \(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)
\(\Leftrightarrow-20x+10x^2-5x^2-10x-5x^2-15x=0\)
\(\Leftrightarrow-45x=0\)
hay x=0