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

\(\left(x+\frac{1}{3}\right)^3=x^3+3.x^2.\frac{1}{3}+3.x.\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\)

\(\left(3x-2y\right)^2=\left(3x\right)^2-2.3x.2y+\left(2y\right)^2=9x^2-12xy+4y^2\)

\(\left(3x-2y\right)^2=\left(3x\right)^2-2.3x.2y+\left(2y\right)^2\)

hđt hả

\(-\frac{32^{xy}}{16^{xy}}=-\frac{32^2}{16^2}=-\left(\frac{32}{16}\right)^2=-\left(2\right)^2=-4\)

\(xy-2x=0\)

\(x\left(y-2\right)=0\)

\(\Rightarrow x=0\) hoặc \(\left(y-2\right)=0\Rightarrow y=2\)

@phynit bài này giải đúng k thầy

thiếu đề x?

áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{3+13}=\frac{2x}{16}=\frac{x}{8}\)

 ta có : \(\frac{x}{8}=\frac{25x}{200}=\frac{xy}{200}\)

\(\Rightarrow25x=xy\)

\(\Rightarrow y=25\)

ta lại có : 

\(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y-\left(x+y\right)}{3-13}=\frac{x-y-x-y}{-10}=\frac{-2y}{-10}=\frac{y}{5}\)

\(\frac{x}{8}=\frac{y}{5}=\frac{x}{8}=\frac{25}{5}=5\)

\(x=5.8=40\)

vậy x=40 ;y=25