(2x)2+2.2x.3+32+x2+2.x.1+12=3x2-6x
<=>4x2+12x+9+x2+2x+1-3x2+6x=0
<=>2x2+20x+10=0
<=>2(x2+10x+5)=0
<=>x2+10x+5=0
<=>x2+2.x.5+52-20=0
<=>(x+5)2-20=0
<=>x+5=+_√20
<=>x=+_√20 -5
Ta có: \(\left(2x+3\right)^2+\left(x+1\right)^2=3x\left(x-2\right)\)
⇌ \(4x^2+12x+9+x^2+2x+1=3x^2-6x\)
⇌ \(2x^2+20x+10=0\)
⇌\(x^2+10x+5=0\)
⇌\(\left(x+5\right)^2-20=0\)
⇌\(\left(x+5+\sqrt{20}\right)\left(x+5-\sqrt{20}\right)=0\)
⇌\(\left[{}\begin{matrix}x+5+\sqrt{20}=0\\x+5-\sqrt{20}=0\end{matrix}\right.\)⇌\(\left[{}\begin{matrix}x=-5-\sqrt{20}\\x=\sqrt{20}-5\end{matrix}\right.\)
Vậy x ∈ \(\left\{-5-\sqrt{20},\sqrt{20}-5\right\}\)
