ĐK: \(x\ge-\dfrac{8}{3}\)
\(\sqrt{x^2+4}=\sqrt{3x+8}\\ \Rightarrow x^2+4=3x+8\\ \Rightarrow x^2-3x-4=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\left(TM\right)\)
\(đkx\ge-\dfrac{8}{3}\\ \left(\sqrt{x^2+4}\right)^2=\left(\sqrt{3x+8}\right)^2\\ x^2+4=3x+8\\ x^2-3x+4-8=0\\ x^2-3x-4=0\\ =>\left[{}\begin{matrix}x=4\left(thoaman\right)\\x=-1\left(thoaman\right)\end{matrix}\right.\)
Vậy ...
ĐK: 3x + 8 >= 0 <=> x>= -8/3
Pt <=> x^2 + 4= 3x + 8
<=> x^2 - 3x -4= 0
<=> x (x+1) - 4 (x +1)=0
<=> (x-4) (x+1)=0
😆<=> (x-4)=0 hoặc (x+1) =0
<=> x= 4 hoặc x= -1 (đều thỏa mãn)
Vậy S= {-1;4}
