Question

3x^2+21x+18+2. căn(x^2+7x+7) = 2

Answer 1

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

\(\Rightarrow3\left(x^2+7x+7\right)-3+2\sqrt{x^2+7x+7}-2=0\)

\(\Rightarrow3\left(x^2+7x+7\right)+2\sqrt{x^2+7x+7}-5=0\)

Đặt \(a=\sqrt{x^2+7x+7}\left(a\ge0\right)\)ta được pt:

3a² + 2a - 5 = 0 => (a - 1)(3a + 5) = 0 => a = 1(nhận) hoặc a = -5/3 (loại)

\(a=1\Rightarrow\sqrt{x^2+7x+7}=1\Rightarrow x^2+7x+7=1\Rightarrow x^2+7x+6=0\)

\(\Rightarrow\left(x+6\right)\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

                                                          Vậy x = -1, x = -6