Ủng hộ cách khác :3
\(x^2+5x-\sqrt{x^2+5x+4}=-2\)
\(\Leftrightarrow x^2+5x+2=\sqrt{x^2+5x+4}\)
Đặt\(\sqrt{x^2+5x+4}=t\) . Phương trình trở thành :
\(t^2-2=t\)
\(\Leftrightarrow t^2-t-2=0\)
\(\Leftrightarrow\left(t+1\right)\left(t-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-1\\t=2\end{matrix}\right.\)
Với \(t=-1\) :
\(\Leftrightarrow\sqrt{x^2+5x+4}=-1\)
\(\Rightarrow\) Phương trình vô nghiệm .
Với \(t=2\) :
\(\Leftrightarrow\sqrt{x^2+5x+4}=2\)
\(\Leftrightarrow x^2+5x=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{-5;0\right\}\)
Wish you study well !!
\(x^2+5x-\sqrt{x^2+5x+4}=-2\)
⇔ \(x^2+5x+2=\sqrt{x^2+5x+4}\)
Ta có : \(VT=x^2+5x+2=x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+2-\dfrac{25}{4}=\left(x+\dfrac{5}{2}\right)^2-\dfrac{17}{4}\text{≥}-\dfrac{17}{4}\left("="\text{⇔ }x=-\dfrac{5}{2}\right)\left(1\right)\) \(VP=\sqrt{x^2+5x+4}=\sqrt{x^2+2.\dfrac{5}{2}x+\dfrac{25}{4}+4-\dfrac{25}{4}}=\sqrt{\left(x+\dfrac{5}{2}\right)^2-\dfrac{9}{4}}\text{≥}-\dfrac{3}{2}\left("="x=-\dfrac{5}{2}\right)\left(2\right)\)
Từ ( 1 ; 2 ) , phương trình vô nghiệm .
