CÂU b nè bạn
\(ĐKXĐ:x\ge3\)
Ta có
\(x^2+8=4\sqrt{x+3}-x\)
<=>\(x^2+8-4\sqrt{x+3}-x=0\)
<=>\(x+3-4\sqrt{x+3}+4+x^2-2x+1\)
<=>\(\left(\sqrt{x-3}-2\right)^2+\left(x-1\right)^2=0\)
<=>
a)\(\sqrt{2011-x^2}-y+x-\sqrt{2011-y^2}=0\)
\(\frac{2011-x^2-y^2}{\sqrt{2011-x^2}+y}+\frac{x^2-2011+y^2}{x+\sqrt{2011-y^2}}=0\)
\(\Leftrightarrow x^2+y^2=2011\)
b) \(x^2-x+4\left(2-\sqrt{x+3}\right)=0\)
\(x\left(x-1\right)+4.\frac{4-x-3}{\sqrt{x+3}+2}=0\)
\(\left(x-1\right)\left(x-\frac{4}{\sqrt{x+3}+2}\right)=0\)
x =1
\(2x+x\sqrt{x+3}-4=0\Leftrightarrow4\left(x-2\right)^2=x^2\left(x+3\right)\)................
