\(x\left(x-y\right)=\frac{3}{10};y\left(x-y\right)=-\frac{3}{50}\)
\(\Rightarrow\frac{x\left(x-y\right)}{y\left(x-y\right)}=\frac{\frac{3}{10}}{\frac{-3}{50}}\Rightarrow\frac{x}{y}=\frac{3}{10}.\frac{50}{-3}=-5\)
\(\Rightarrow x=-5y\)(1)
Thay (1) vào
\(x\left(x-y\right)=\frac{3}{10}\Rightarrow-5y\left(-5y-y\right)=\frac{3}{10}\)
\(\Rightarrow30y^2=\frac{3}{10}\Leftrightarrow y^2=\frac{3}{10}.\frac{1}{30}=\frac{1}{100}\Rightarrow y=\orbr{\begin{cases}\frac{1}{10}\\-\frac{1}{10}\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{1}{10}.-5=-\frac{1}{2}\\-\frac{1}{10}.-5=\frac{1}{2}\end{cases}}\)
x*(x-y)=3/10
y*(x-y)=-3/50
=>x*(x-y)-y*(x-y)=3/10-(-3/50)
=>(x-y)*(x-y)=15/50+3/50=18/50=9/25
=>x-y=(3/5;-3/5)
+)Nếu x-y=3/5=>x=1/2;y=-10
+)Nếu x-y=-3/5=>x=-1/2;y=10
Phạm Tuấn Đạt ơi sao suy luôn ra đc 30y^2 z ?