x+y=x;y=3.[x-y]
x+y=x.\(\frac{1}{y}\)=3[x-y]
-x-y=x.\(\frac{1}{y}\)=3.[x-y]
=x.\(\frac{1}{y}\)=3[x-y]+x-y
=x\(\frac{1}{y}\)=4 .[x-y]
=1/yx=4x-4y
=1/yx-4x=4y
=1=4y
y=1/4
Có: \(x+y=3\left(x-y\right)\)\(x+y=3x-3y\)\(2x=4y\)\(\frac{x}{y}=2\)\(x=2y\)
Thay x = 2y vào x + y, ta được: \(3y=\frac{x}{y}\)
Có: \(\hept{\begin{cases}\frac{x}{y}=3y\\\frac{x}{y}=2\end{cases}\Rightarrow3y=2\Rightarrow y=\frac{2}{3}}\)
\(\Rightarrow x=2.\frac{2}{3}=\frac{4}{3}\)
Vậy \(x=\frac{4}{3};y=\frac{2}{3}\)
Dòng đầu: \(x+y=3\left(x-y\right)\)
\(x+y=3x-3y\)
\(2x=4y\)
\(\frac{x}{y}=2\)
\(x=2y\)
\(x+y=x:y=3\left(x-y\right)\)\(\left(Đk:y\ne0\right)\)
Ta có: \(x+y=3\left(x-y\right)\)
\(\Leftrightarrow x+y=3x-3y\)
\(\Leftrightarrow2x=4y\)
\(\Leftrightarrow x=2y\)
\(\Rightarrow x:y=2\)
Ta lại có:\(x+y=x:y\)
\(\Rightarrow x+y=2\)
Mà\(x=2y\)
\(\Rightarrow2y+y=2\)
\(\Leftrightarrow3y=2\)
\(\Leftrightarrow y=\frac{2}{3}\)(Thỏa mãn Đk:\(y\ne0\))
Mà \(x=2y\)
\(\Rightarrow x=2.\frac{2}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Vậy\(x=\frac{4}{3};y=\frac{2}{3}\)
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