căng nha

Giúp mình câu này với ah. 

a Ah box môn TA đâu gòi:)?

X³ + Y³ = X³ + 3X²Y + 3 XY²+ Y²+ 3XY - 3 X²Y- 3XY²

=(x + y )³  + 3xy. ( 1 - x - y )

=( x + y)³ + 3xy . [ 1 - (x - y) ]

= 1³ + 3xy. ( 1-1)

=1

mik cũng ko chắc nữa nhé

Ta có :x³ +y³ +3xy=(x+y)(x² -xy+y²)+3xy

mà x+y=1

=>x² -xy+y²+3xy=x² +2xy+y² =(x+y)²=1² =1

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

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

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

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