Giải PT(1) [tex]\sqrt{x}+2\sqrt{x+3}=7-\sqrt{x^2+3}[/tex]
[tex]\Leftrightarrow \sqrt{x}-1+2(\sqrt{x+3}-2)+\sqrt{x^2+3}-2=0[/tex]
[tex]\Leftrightarrow \frac{x-1}{\sqrt{x}+1}+2\frac{x-1}{\sqrt{x+3}+2}+\frac{x^2-1}{\sqrt{x^2+3}+2}=0[/tex]
[tex]\Leftrightarrow (x-1)(\frac{1}{\sqrt{x}+1}+\frac{2}{\sqrt{x+3}+2}+\frac{x +1}{\sqrt{x^2+3}+2})=0[/tex]
Do ĐK:[tex]x\geq 0\Rightarrow[/tex] [tex]\frac{1}{\sqrt{x}+1}+\frac{2}{\sqrt{x+3}+2}+\frac{x +1}{\sqrt{x^2+3}+2}\neq 0[/tex]
[tex]\Leftrightarrow x=1[/tex]
Thay vào Pt(2) rồi giải.
[tex]\sqrt{y+1}+\sqrt{7-y}=y^2-6y+13[/tex]
[tex]\Leftrightarrow \sqrt{y+1}-2+\sqrt{7-y}-2=y^2-6y+9=(y-3)^2[/tex]
Giải như trên.