\(a.3x^2+21x+18+2\sqrt{x^2+7x+7}=2\)
\(\Leftrightarrow3\left(x^2+7x+6\right)+2\sqrt{x^2+7x+7}=2\circledast\)
Đặt : \(x^2+7x+7=t\left(t\ge0\right)\) , ta có :
\(\circledast\Leftrightarrow3\left(t-1\right)+2\sqrt{t}=2\)
\(\Leftrightarrow3t+2\sqrt{t}-5=0\)
\(\Leftrightarrow3\sqrt{t}\left(\sqrt{t}-1\right)+5\left(\sqrt{t}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-1=0\\3\sqrt{t}+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(TM\right)\\vô-nghiệm\end{matrix}\right.\)
Với : \(t=1\) , thì : \(x^2+7x+7=1\Leftrightarrow x^2+x+6x+6=0\)
\(\Leftrightarrow x\left(x+1\right)+6\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
KL--...
\(b.2x^2-8x-3\sqrt{x^2-4x-5}=12\circledast\)
ĐKXĐ : \(\left[{}\begin{matrix}x\ge5\\x\le-1\end{matrix}\right.\)
\(\circledast\Leftrightarrow2x^2-8x-12-3\sqrt{x^2-4x-5}=0\)
\(\Leftrightarrow2\left(x^2-4x-3\right)-3\sqrt{x^2-4x-5}=0\)
Đặt : \(x^2-4x-5=t\left(t\ge0\right)\) , ta có :
\(2\left(t+2\right)-3\sqrt{t}=0\)
\(\Leftrightarrow2t-3\sqrt{t}+4=0\)
\(\Leftrightarrow2\left(t-2.\dfrac{3}{4}\sqrt{t}+\dfrac{9}{16}\right)+4-\dfrac{9}{8}=0\)
\(\Leftrightarrow\left(\sqrt{t}-\dfrac{3}{4}\right)^2=\dfrac{23}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{t}-\dfrac{3}{4}=\dfrac{\sqrt{23}}{4}\\\sqrt{t}-\dfrac{3}{4}=-\dfrac{\sqrt{23}}{4}\end{matrix}\right.\)
Tới đây dễ rồi , bạn tự làm nốt nhé...:)
