\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
\(\Leftrightarrow x^2+5x+4-3\sqrt{x^2+5x+2}=6\)
Đặt : \(x^2+5x+2=t\left(t\ge0\right)\) , ta có :
\(t+2-3\sqrt{t}=6\)
\(\Leftrightarrow t-3\sqrt{t}-4=0\)
\(\Leftrightarrow\left(\sqrt{t}+1\right)\left(\sqrt{t}-4\right)=0\)
\(\Leftrightarrow t=16\left(TM\right)\)
Ta có : \(x^2+5x+2=16\)
\(\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
KL...........
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