Question

Answer 1

\(\sqrt{x}+2\sqrt{x+3}=7-\sqrt{x^2+3}\)\(\left(đk:x\ge0\right)\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2\left(x-1\right)}{\sqrt{x+3}+2}+\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+3}+2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x+3}+2}+\dfrac{x+1}{\sqrt{x^2+3}+2}\right)=0\)

Vì \(x\ge0\Rightarrow\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{x+3}+2}+\dfrac{x+1}{\sqrt{x^2+3}+2}>0\)

=> x=1 thay vào pt 2

\(\Leftrightarrow\sqrt{3+6}+\sqrt{1+y-4}=5\)

\(\Leftrightarrow\sqrt{y-3}=2\) (\(Đk:y\ge3\))

\(\Leftrightarrow y-3=4\Rightarrow y=7\)

Vậy \(\left(x;y\right)=\left(1,7\right)\)