What ??!!!

Có x+y=3 rồi tính làm gì nữa

Bạn kia ns đúng á tại mình bấm lộn nút, sorry nha Lê Tài Bảo Châu

y²-(y+x)+xy+x²+3

min P=x^2+y^2+xy-3(x+y)+3

a: \(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{y}+\sqrt{x}}=\frac{x+y}{\sqrt{x}+\sqrt{y}}\)

Ta có: \(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(=\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}-\frac{\left(x+y\right)_{}\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(\) \(=\frac{x^2-\sqrt{xy}\left(x+y\right)-y^2-x^2+y^2}{\sqrt{xy}\left(x-y\right)}=\frac{-\left(x+y\right)}{x-y}\)

b: Thay x=3; \(y=4+2\sqrt3\) vào A, ta được:

\(A=\frac{-\left(3+4+2\sqrt3\right)}{3-\left(4+2\sqrt3\right)}=\frac{-7-2\sqrt3}{-2\sqrt3-1}=\frac{7+2\sqrt3}{2\sqrt3+1}\)

\(=\frac{\left(7+2\sqrt3\right)\left(2\sqrt3-1\right)}{12-1}=\frac{14\sqrt3-7+12-2\sqrt3}{11}=\frac{12\sqrt3+5}{11}\)

Ta có: \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac52\)

=>\(\frac{x}{\sqrt{xy}}+\frac{y}{\sqrt{xy}}=\frac52\)

=>\(2\left(x+y\right)=5\sqrt{xy}\)

=>\(2x-5\sqrt{xy}+2y=0\)

=>\(2x-4\sqrt{xy}-\sqrt{xy}+2y=0\)

=>\(2\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)=0\)

=>\(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{x}-\sqrt{y}\right)=0\)

TH1: \(\sqrt{x}-2\sqrt{y}=0\)

=>\(\sqrt{x}=2\sqrt{y}\)

=>x=4y

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)

\(=\frac{2\cdot\sqrt{4y}+3\sqrt{y}}{2\cdot\sqrt{4y}-3\sqrt{y}}=\frac{4\sqrt{y}+3\sqrt{y}}{4\sqrt{y}-3\sqrt{y}}=\frac{4+3}{4-3}=7\)

TH2: \(2\sqrt{x}-\sqrt{y}=0\)

=>\(\sqrt{y}=2\sqrt{x}\)

=>y=4x

\(A=\frac{2x+3\cdot\sqrt{xy}}{2x-3\sqrt{xy}}\)

\(=\frac{\sqrt{x}\left(2\sqrt{x}+3\sqrt{y}\right)}{\sqrt{x}\left(2\sqrt{x}-3\sqrt{y}\right)}=\frac{2\sqrt{x}+3\sqrt{y}}{2\sqrt{x}-3\sqrt{y}}\)

\(=\frac{2\sqrt{x}+3\cdot\sqrt{4x}}{2\sqrt{x}-3\cdot\sqrt{4x}}=\frac{2\sqrt{x}+6\sqrt{x}}{2\sqrt{x}-6\sqrt{x}}=\frac{8}{-4}=-2\)

có: \(\dfrac{1}{x^2+y^2}=\dfrac{1}{\left(x+y\right)^2-2xy}=\dfrac{1}{1-2xy}\)(1)

có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)

từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)

=>Min A=(1+\(\sqrt{2}\))^2

cảm ơn rất nhiều