Question

Giải PT: \(3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)

Answer 1

\(PT\Leftrightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0.\)

\(\Leftrightarrow3\left(x^2+7x+7\right)+2\sqrt{x^2+7x+7}-5=0\)

Đặt \(a=\sqrt{x^2+7x+7}\)(a\(\ge\)0)

\(PT\Leftrightarrow3a^2+2a-5=0\)

\(\Leftrightarrow\left(a-1\right)\left(3a+5\right)=0\)

Vì a\(\ge\)0 nên a-1=0=> a=1

lúc đó x²+7x+7=1

<=> x²+7x+6=0

<=> (x+1)(x+6)=0

<=> \(\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Vậy.................................