Dk: x≥-5
Đặt √(x+5)=t (t≥0)
{x²-5=t(1)
{x+5=t²<=>t²-5=x(2)
lay (1)-(2):
(x-t)(x+t+1)=
giai tiep
ta có đk x^2-9>0=>x<-3;x>3
\(\frac{3x}{\sqrt{x^2-9}}+x-3\sqrt{2^3}=0\)
<=>\(\frac{\left(x-3\sqrt{2^3}\right)\sqrt{x^2-9}+3x}{\sqrt{x^2-9}}=0\)
\(0=\frac{1}{\sqrt{x^2-9}}\)\(\left(x-3\sqrt{2^3}\right)\sqrt{x^2-9}+3x=0\)=>\(x^4-3\sqrt{2^5}x^3+54x^2+27\sqrt{2^5}x-648=0\)
=>x=3\(\sqrt{2}\)