Áp dụng t/c dtsbn:

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{2x}{14}=\dfrac{2x-y}{14-6}=\dfrac{120}{8}=15\)

\(\Rightarrow\left\{{}\begin{matrix}x=15.7=105\\y=15.6=90\end{matrix}\right.\)

\(x:y=7:6\Leftrightarrow\frac{x}{7}=\frac{y}{6}=\frac{x+y}{7+6}=\frac{39}{13}=3\)

\(\Rightarrow\frac{x}{7}=3\Leftrightarrow x=21;\frac{y}{6}=3\Leftrightarrow y=18\)

...

ta có : \(x:y=7:6\Rightarrow\frac{x}{7}=\frac{y}{6}\)

áp dụng tính chất dãy tỉ số bằng nhau, ta có :

\(\frac{x}{7}=\frac{y}{6}=\frac{2x}{14}=\frac{y}{6}=\frac{2x-y}{14-6}=\frac{120}{8}=15\)

\(\frac{x}{7}=15\rightarrow x=105\)

\(\frac{y}{6}=15\rightarrow y=90\)

A

A

A. x = 105, y = 90  

A

a

Lời giải:

Đặt $\frac{x}{3}=\frac{y}{2}=t$

$\Rightarrow x=3t; y=2t$. Thay vô điều kiện $4x-y=20$ ta có:

$4.3t-2t=20$

$\Leftrightarrow 10t=20\Leftrightarrow t=2$
$\Rightarrow x=3t=6; y=2t=4$

Ta có: \(\frac{x}{7}=\frac{y}{6}=\frac{2x}{14}=\frac{y}{6}=\frac{2x-y}{14-6}=\frac{120}{8}=15\)

\(\Rightarrow\frac{x}{7}=15\Rightarrow x=105;\frac{y}{6}=15\Rightarrow y=90\)

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

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

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

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