Lời giải:

 $\frac{x}{y}=\frac{2}{3}\Rightarrow \frac{x}{2}=\frac{y}{3}$. Đặt $\frac{x}{2}=\frac{y}{3}=k$ thì:

$x=2k; y=3k$

Khi đó: $3x-2y=3.2k-3.2k=0$. Mẫu số không thể bằng $0$ nên $A$ không xác định. Bạn xem lại.

$B=\frac{2(2k)^2-2k.3k+3(3k)^2}{3(2k)^2+2.2k.3k+(3k)^2}=\frac{29k^2}{33k^2}=\frac{29}{33}$

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow12x-4y-3y=3x\)

\(\Rightarrow12x-7y=3x\)

\(\Rightarrow12x-3x=7y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Ta có:

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow12x-4y=3x+3y\\ \Rightarrow9x=7y\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy.........

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\Rightarrow\left(3x-y\right).4=3\left(x+y\right)\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\)\(\dfrac{x}{y}=\dfrac{7}{9}\)

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\Rightarrow4\left(2x-y\right)=3\left(x+y\right)\)

\(\Rightarrow12x-4y=3x+3y\Rightarrow12x-3x=3y+4y\)

\(\Rightarrow9x=7y\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

sửa lại cho mik dòng đầu là 4(3x-y)

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Leftrightarrow12x-4y=3x+3y\)

\(\Leftrightarrow12x=3x+7y\)

\(\Leftrightarrow9x=7y\)

\(\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{9}\Leftrightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Ta có : \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Leftrightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Leftrightarrow12x-4y=3x+3y\Rightarrow15x=7y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{15}\)

tik mik nha !!!

Xét \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)

\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)

Xét \(x+y+z\ne0\) thì ta có:

\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)

\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)

Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)

Nếu bị lỗi thì bạn có thể xem đây nhé:

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\\ \Rightarrow4\cdot\left(3x-y\right)=3\cdot\left(x+y\right)\\ \Leftrightarrow12x-4y=3x+3y\\ \Leftrightarrow12x-3x=3y+4y\\ \Leftrightarrow9x=7y\\ \Rightarrow\dfrac{x}{7}=\dfrac{y}{9}\\ \Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy \(\dfrac{x}{y}=\dfrac{7}{9}\)

Ta có: \(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow\left(3x-y\right)4=\left(x+y\right)3\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow12x-3x=4y+3y\)

\(\Rightarrow9x=7y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

Vậy \(\dfrac{x}{y}=\dfrac{7}{9}.\)

\(\dfrac{3x-y}{x+y}=\dfrac{3}{4}\)

\(\Rightarrow4\left(3x-y\right)=3\left(x+y\right)\)

\(\Rightarrow12x-4y=3x+3y\)

\(\Rightarrow12x-4y-3y=3x\)

\(\Rightarrow12x-7y=3x\)

\(\Rightarrow9x=7y\)

\(\Rightarrow x=\dfrac{7}{9}y\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{7}{9}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)