dẽ quá luôn

tự làm nha

siêu dễ luôn

\(\hept{\begin{cases}x\left(x-y\right)=\frac{3}{10}\\y\left(x-y\right)=-\frac{3}{50}\end{cases}}\Rightarrow\frac{x}{y}=-5\Rightarrow x=-5y\)

\(x\left(x-y\right)=-5y\left(-5y-y\right)=30y^2=\frac{3}{10}\Leftrightarrow\orbr{\begin{cases}y=\frac{1}{10}\Rightarrow x=-\frac{1}{2}\\y=-\frac{1}{10}\Rightarrow x=\frac{1}{2}\end{cases}}\)

\(x\left(x-y\right)-y\left(x-y\right)=\frac{3}{10}+\frac{3}{50}=\frac{18}{50}=\frac{9}{25}\)

\(\Rightarrow\left(x-y\right)\left(x-y\right)=\frac{9}{25}\)

\(\Rightarrow\left(x-y\right)^2=\orbr{\begin{cases}\left(\frac{3}{5}\right)^2\\\left(-\frac{3}{5}\right)^2\end{cases}}\)

\(\Rightarrow x-y=\orbr{\begin{cases}\frac{3}{5}\\-\frac{3}{5}\end{cases}}\)

\(\frac{x.\left(x-y\right)}{y.\left(x-y\right)}=\frac{3}{10}:-\frac{3}{50}=-5=\frac{x}{y}\)

\(x=-5y\Rightarrow-5y\left(-5y-y\right)=\frac{3}{10}\Rightarrow30y^2=\frac{3}{10}\Rightarrow y=\frac{1}{10}hoặcy=-\frac{1}{10}\)

với y=1/10 => x=-5.1/10=-1/2

với y= -1/10 => x= 1/2

câu trả lời đúng ko đấy

x/8=ý/10và x+ý=72

\(x\left(x-y\right)=\dfrac{3}{10}\)

\(y\left(x-y\right)=-\dfrac{3}{50}\)

=>\(\dfrac{x\left(x-y\right)}{y\left(x-y\right)}=\dfrac{3}{10}:\dfrac{-3}{50}=-5\)

=>x/y=-5

=>x=-5y

x(x-y)=3/10

=>\(-5y\left(-5y-y\right)=\dfrac{3}{10}\)

=>\(5y\cdot6y=\dfrac{3}{10}\)

=>\(30y^2=\dfrac{3}{10}\)

=>\(y^2=\dfrac{1}{100}\)

=>\(\left[{}\begin{matrix}y=\dfrac{1}{10}\\y=-\dfrac{1}{10}\end{matrix}\right.\)

TH1: \(y=\dfrac{1}{10}\)

=>\(x=-5y=-5\cdot\dfrac{1}{10}=-\dfrac{5}{10}=-\dfrac{1}{2}\)

TH2: y=-1/10

=>\(x=-5y=-5\cdot\dfrac{-1}{10}=\dfrac{1}{2}\)

Từ gt,ta có : x(x - y) - y(x - y) =\(\frac{3}{10}-\frac{-3}{50}\)

(x - y)² =\(\frac{9}{25}\)\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}:\frac{3}{5}=\frac{1}{2}\\x=\frac{3}{10}:\frac{-3}{5}=\frac{-1}{2}\end{cases};\orbr{\begin{cases}y=\frac{-3}{50}:\frac{3}{5}=\frac{-1}{10}\\y=\frac{-3}{50}:\frac{-3}{5}=\frac{1}{10}\end{cases}}}}\)

Vậy\(x=\frac{1}{2};y=\frac{-1}{10}\) hoặc\(x=\frac{-1}{2};y=\frac{1}{10}\)

mình cũng làm giống bạn  

b/

Ta có \(\left(x-3\right)\left(x+\frac{1}{2}\right)>0\)

=> \(\orbr{\begin{cases}x-3>0\\x+\frac{1}{2}>0\end{cases}}\)=> \(\orbr{\begin{cases}x>3\\x>\frac{-1}{2}\end{cases}}\)