Question

ghpt

1) \(\left\{{}\begin{matrix}3\left(2-x\right)\sqrt{2-y^2}=2-y+\dfrac{4}{x+1}\\\left(x^2+xy-x+y-2\right)\sqrt{2-y^2}+2=x+y\end{matrix}\right.\)

Answer 1

part full :v

*Th 1: \(x+y=2\)

\(Pt\left(1\right)\Leftrightarrow3y\sqrt{2-y^2}=x+\dfrac{4}{x+1}\)

xét \(VT=3y\sqrt{2-y^2}=3\sqrt{y^2\left(2-y^2\right)}\le3.\dfrac{y^2+2-y^2}{2}=3\)(theo AM-GM)

\(VT=x+\dfrac{4}{x+1}=\left(x+1\right)+\dfrac{4}{x+1}-1\ge2\sqrt{\dfrac{4\left(x+1\right)}{x+1}}-1=4-1=3\)(theo AM-GM)

do đó \(VT\le3;VF\ge3\)

\(VT=VF\Leftrightarrow\left\{{}\begin{matrix}y^2=2-y^2\\x+1=\dfrac{4}{x+1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\pm1\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\)(tmđkxđ)(4 cặp)

*TH 2 \(\left(x+1\right)\sqrt{2-y^2}=1\Leftrightarrow x+1=\dfrac{1}{\sqrt{2-y^2}}\)(\(-\sqrt{2}< y< \sqrt{2}\))

thế vào Pt(1) , bình phương giải (nhác làm quá)

Answer 2

\(Pt\left(2\right)\Leftrightarrow\left(x+y-2\right)\left[\left(x+1\right)\sqrt{2-y^2}-1\right]=0\)