a) $DK:..$
Ta có: [tex]\sqrt{x+8}+2\sqrt{x+5}\geq 0[/tex]
Dấu ''='' xảy ra khi: [tex]\left\{\begin{matrix} x+5=0 & & \\ x+8=0 & & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-5 & & \\ x=-8 & & \end{matrix}\right.\Rightarrow False[/tex]
Vậy $PTVN_0$
b) Đặt: [tex]\sqrt{x+7}=a;\sqrt{9x+56}=b(a;b\geq 0)[/tex]
Khi đó: $9a^2-b^2=7$
Khi đó ta có $HPT$
[tex]\left\{\begin{matrix} a-b=12 & & \\ 9a^2-b^2=7 & & \end{matrix}\right.\Rightarrow a;b=..\Rightarrow x=...[/tex]