Vì \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\) nên \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}\)

\(\Rightarrow\)\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)

Vậy giá trị của mỗi tỉ số đó bằng \(\dfrac{1}{2}\)

1)

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=k\left(k\in Q\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\end{matrix}\right.\)

Vì \(xy=90\) nên \(2k.5k=90\)

\(\Rightarrow10k^2=90\)

\(\Rightarrow k^2=9\)

\(\Rightarrow\left[{}\begin{matrix}k=3\\k=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=15\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-15\end{matrix}\right.\end{matrix}\right.\)

Vậy có 2 cặp số (x;y) thảo mãn là: (6; 15); (-6; -15)

1) Giải

Đặt : \(\dfrac{x}{2}=\dfrac{y}{5}=k\Rightarrow x=2k;y=5k\)

Thay vào xy=90

Ta có :2k\(\cdot\)5k= 90

\(\Rightarrow\) \(10\cdot k^2=90\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k^2=3^2\)

\(\Rightarrow k=3\)

Vậy x=\(2\cdot3\)=6;y=\(5\cdot3=15\)

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

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

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

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

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

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

\(\Rightarrow9x=7y\)

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

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

Ta có:

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

Vậy.........

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

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

\(\Rightarrow9x=7y\)

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

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.

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

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

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

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

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 !!!

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

=> 4(3x-y)=3(x+y)

=> 12x-4y=3x+3y

=> 12x-3x=4y+3y

=> 9x=7y

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

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

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

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

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

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

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

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

\(\Rightarrow9x=7y\)

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

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