\(1,\sqrt{x^2+5x+20}=4\)
\(\Leftrightarrow\sqrt{\left(x^2+5x+20\right)^2}=4^2\)
\(\Leftrightarrow x^2+5x+20=16\)
\(\Leftrightarrow x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x\left(x+1\right)+4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{-4;-1\right\}\)
\(2,3-\sqrt{x^2+5}=4\)
\(\Leftrightarrow\sqrt{x^2+5}=-1\)
\(\Leftrightarrow\sqrt{\left(x^2+5\right)^2}=\left(-1\right)^2\)
\(\Leftrightarrow x^2+5=1\)
\(\Leftrightarrow x^2=-4\) (loại)
Vậy \(S=\varnothing\)
Lời giải:
1. ĐKXĐ:...........
PT $\Leftrightarrow x^2+5x+20=4^2=16$
$\Leftrightarrow x^2+5x+4=0$
$\Leftrightarrow (x+1)(x+4)=0$
$\Leftrightarrow x+1=0$ hoặc $x+4=0$
$\Leftrightarrow x=-1$ hoặc $x=-4$ (thỏa mãn)
Vậy...............
2.ĐKXĐ:.........
PT $\Leftrightarrow \sqrt{x^2+5}=3-4=-1<0$ (vô lý)
Vậy pt vô nghiệm.
