Ta có:\(\orbr{\begin{cases}2x-3y=3\\x+2y=2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}4x-6y=6\\3x+6y=6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}7x=12\\3x+6y=6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{12}{7}\\3x+6y=6\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{12}{7}\\y=\frac{1}{7}\end{cases}}\)

Vậy tỉ lệ thức \(\frac{y}{x}=\frac{1}{12}\)

Có \(\frac{3x-y}{x+y}=\frac{1}{2}\)

=> (3x- y).2 = x + y

=> 6x - 2y = x + y 

=> 5x = 3y

=> \(\frac{x}{y}=\frac{3}{5}\)

\(\frac{3x-y}{x+y}=\frac{1}{2}\)

\(\Rightarrow2\left(3x-y\right)=x+y\)

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

\(\Rightarrow5x=3y\)

\(\Rightarrow\frac{x}{y}=\frac{3}{5}\)

Vậy :................

Câu a:

0,4 : x = x : 0,9

x^2 = 0,4 x 0,9

x^2 = 0,36

x^2 = (0,6)^2

x = - 0,6 hoặc x = 0,6

Vậy x ∈ {-0,6; 0,6}

Câu b:

13\(\frac13\) : 1\(\frac13\) = 26 : (2x - 1)

\(\frac{40}{13}\) : \(\frac43\) = 26 :(2x - 1)

\(\frac{40}{13}\) x \(\frac34\) = 26 :(2x - 1)

\(\frac{30}{13}\) = 26 : (2x - 1)

26 : \(\frac{30}{13}\) = 2x - 1

26 x \(\frac{13}{30}\) = 2x - 1

169/15 = 2x - 1

2x = 169/15 + 1

2x = 184/15

x = 184/15 : 2

x = 184/15 x 1/2

x = 92/15

Vậy x = 92/15

Ta có:

\(\left(2x-3y\right).3=\left(x+2y\right).2\)

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

\(6x-9y=2x+4y\)

\(6x-2x=9y+4y\)

\(4x=13y\)

Chia hai vế cho 4y, ta có:

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

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

\(\Rightarrow\) Tỉ số giữa x và y là \(\frac{13}{4}\)

Ta có:

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

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

<=> 6x - 9y = 2x + 4y

<=> 6x - 2x = 9y+4y

<=> 4x = 13y

=> \(\frac{x}{y}=\frac{13}{4}\)

Giải:
Ta có: \(\frac{x+2}{y+3}=\frac{2}{3}\Rightarrow3\left(x+2\right)=2\left(y+3\right)\)

\(\Rightarrow3x+6=2y+6\)

\(\Rightarrow3x=2y\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=k\)

\(\Rightarrow x=2k,y=3k\)

Lại có: \(A=\frac{x^2+y^2}{xy}=\frac{\left(2k\right)^2+\left(3k\right)^2}{2k3k}=\frac{4k^2+9k^2}{6k^2}=\frac{\left(4+9\right)k^2}{6k^2}=\frac{13}{6}\)

Vậy \(A=\frac{13}{6}\)

\(A=\frac{13}{6}\)

\(\frac{3x-y}{x+y}=\frac{1}{2}\)

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

=> 6x - 2y = x + y

=> 5x = 3y

=> \(\frac{x}{y}=\frac{3}{5}\)