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\(\frac{2x-y}{x+y}=\frac{2}{3}\Rightarrow2\left(x+y\right)=3\left(2x-y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3.\left(2x-y\right)=2.\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=3y+2y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy tỉ số \(\frac{x}{y}=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo bài ra ta có:
2x-y/x+y=2/3
=>3(2x-y)=2(x+y)
6x-3y=2x+2y
=>(6x-3y)-(2x+2y)=0
6x-3y-2x-2y=0
4x-y=0
=>4x=y
=>x/y=1/4
Theo đầu bài ta có:
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x-y}{x+y}=\frac{2}{3}\Rightarrow\frac{2x-y}{2}=\frac{x+y}{3}=\frac{\left(2x-y\right)-\left(x+y\right)}{2-3}=2y-x\)
\(\Rightarrow2x-y=4y-2x\Rightarrow4x=5y\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Áp dụng công thức lớp 7 ; \(\frac{a}{b}\)= \(\frac{c}{d}\) thì \(\frac{a}{c}\)= \(\frac{b}{d}\)
thì \(\frac{2x-y}{2}\)= \(\frac{x+y}{3}\)= \(\frac{2x-y-\left(x+y\right)}{2-3}\)= \(\frac{x-2y}{-1}\)= - (x - 2y ) = - x + 2y = 2y + (- x) = 2y - x
=> .....................................x/y = 5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)
=> (2x - y).3 = (x+y) .2
6x - 3y = 2x + 2y
6x - 2x = 3y + 2y
4x = 5y
=> \(\frac{x}{5}\)=\(\frac{y}{4}\)
Vậy tỉ số \(\frac{x}{y}\)=\(\frac{5}{4}\)
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy \(\frac{x}{y}=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{2x-y}{x+y}=\frac{2}{3}\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\Leftrightarrow6x-3y=2x+2y\Leftrightarrow4x=5y\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vì \(\frac{2x-y}{x+y}=\frac{2}{3}=>\left(2x-y\right).3=\left(x+y\right).2=>6x-3y=2x+2y\)
\(=>6x-2x=2y-\left(-3y\right)=>6x-2x=2y+3y=>4x=5y=>\frac{x}{y}=\frac{5}{4}\)
Vậy tỉ số x/y=5/4
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y-2x-2y=0\)
\(\Leftrightarrow4x-5y=0\)
\(\Leftrightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)
ta có \(\frac{2x-y}{x+y}=\frac{2}{3}\)<=> 3.( 2x-y)= 2.(x+y) <=> 6.x -3.y=2.x+2.y<=> 6.x-2.x=3.y+2.y <=> 4.x=5.y
Suy ra \(\frac{x}{y}=\frac{5}{4}\)