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

\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

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

\(\Rightarrow4x=5y\)

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

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

\(\Rightarrow2x-y=4y-2x\Rightarrow4x=5y\Rightarrow\frac{x}{y}=\frac{5}{4}\)

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

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

Theo bài ra ta có: \(\frac{2x-y}{x+y}=\frac{2}{3}\)

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

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

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

=> 4x=5y

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

Vậy \(\frac{x}{y}=\frac{5}{4}\)

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

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

\(\Rightarrow3\left(2x-y\right)=2\left(x+y\right)\)

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

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

\(\Rightarrow4x=5y\)

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

\(\frac{2x-3y}{x+2y}=\frac{2}{3}=>\left(2x-3y\right).3=\left(x+2y\right).2=>6x-9y=2x+4y=>6x-2x=9y+4y\)

=>4x=13y

hay \(\frac{x}{y}=\frac{13}{4}\)

vậy gtri của tỉ số x/y là 13/4

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

a

Nếu  \(y=0\Rightarrow x^2=3025\Rightarrow x=55\)

Nếu \(y>0\Rightarrow3^y⋮3\)

Mà \(3026\equiv2\left(mod3\right)\Rightarrow x^2\equiv2\left(mod3\right)\) 9 vô lý

Vậy.....

b

Không mất tính tổng quát giả sử \(x\ge y\)

Ta có:

\(\frac{1}{2}=\frac{1}{2x}+\frac{1}{2y}+\frac{1}{xy}\le\frac{1}{2y}+\frac{1}{2y}+\frac{1}{y^2}=\frac{1}{y}+\frac{1}{y^2}=\frac{y+1}{y^2}\)

\(\Rightarrow y^2\le2y+2\Rightarrow\left(y^2-2y+1\right)\le3\Rightarrow\left(y-1\right)^2\le3\Rightarrow y\le2\Rightarrow y=1;y=2\)

Với \(y=1\Rightarrow\frac{1}{2x}+\frac{1}{2}+\frac{1}{x}=\frac{1}{2}\Rightarrow\frac{1}{2x}+\frac{1}{x}=0\) ( loại )

Với \(y=2\Rightarrow\frac{1}{2x}+\frac{1}{4}+\frac{1}{2x}=\frac{1}{2}\Rightarrow\frac{1}{x}=\frac{1}{4}\Rightarrow x=4\)

Vậy x=4;y=2 và các hoán vị

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