Question

Tìm tỉ số \(\frac{x}{y}\)biết \(\frac{2x-y}{x+y}=\frac{2}{3}.\)

Answer 1

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

Answer 2

Theo đầu bài ta có:
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Rightarrow6x-3y=2x+2y\)
\(\Rightarrow6x-2x=2y+3y\)
\(\Rightarrow4x=5y\)
\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)

Answer 3

\(\frac{2x-y}{x+y}=\frac{2}{3}\)\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Rightarrow6x-3y=2x+2y\)

\(\Rightarrow4x=5y\)

\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)

Vậy.........

Answer 4

Cho mình sửa lại:

6x-3y-2x-2y=0

4x-5y=0

4x=5y

=>x/y=5/4

Answer 5

Thử lại:
Do \(\frac{x}{y}=\frac{5}{4}\Rightarrow x=5k;y=4k\)
Từ đó suy ra:\(\frac{2x-y}{x+y}=\frac{2\cdot5k-4k}{5k+4k}=\frac{10k-4k}{5k+4k}=\frac{6k}{9k}=\frac{2\cdot3k}{3\cdot3k}=\frac{2}{3}\)( đpcm )
Vậy \(\frac{x}{y}=\frac{5}{4}\)

Answer 6

Ta có:

\(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

\(\Leftrightarrow6x-3y=2x+2y\)\(\Leftrightarrow6x-2x=2y+3y\)\(\Leftrightarrow4x=5y\)\(\Rightarrow\frac{x}{y}=\frac{5}{4}\)

Answer 7

Ta có:\(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)suy ra:[2x-y]3=[x+y]2

                                    suy ra:6x-3y=2x+2y

                                           =6x-2x=3y+2y

                                           =4x=5y

                                  suy ra:\(\frac{x}{y}\)=\(\frac{4}{5}\)

                Vậy x=4;y=5

Answer 8

Ta có :  \(\frac{2x-y}{x+y}\)=\(\frac{2}{3}\)

=>(2x-y).3= 2(x+y)

      6x-3y = 2x+2y

     6x-2x =3y+2y

     4x=5y

    => 5/4=x/y