Question

Tìm 2 số x và y,biết:
a)\(\dfrac{x}{3}=\dfrac{y}{7}\)và x+y=20 b)\(\dfrac{x}{5}\) =\(\dfrac{y}{2}\) và x-y=6

Answer 1

a/ Ta có :

\(x+y=20\)

\(\dfrac{x}{3}=\dfrac{y}{7}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=2\Leftrightarrow x=6\\\dfrac{y}{7}=2\Leftrightarrow y=14\end{matrix}\right.\)

Vậy ..

b/ Ta có :

\(x-y=6\)

\(\dfrac{x}{5}=\dfrac{y}{2}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=2\Leftrightarrow x=10\\\dfrac{y}{2}=2\Leftrightarrow y=4\end{matrix}\right.\)

Vậy ..

Answer 2

A/d t/c of dtsbn t.có

x/3=y/7=>x+y/3+7=20/10=2

x/3=2=>x=6

y/7=2=>y=14

x/5=y/2=>x-y/5-2=6/3=2

x/5=2=>x=10

y/2=2=>y=4

Answer 3

a,

a)$\frac{x}{3}=\frac{y}{7}$và x+y=20

=> \(\dfrac{x+y}{3+7}=\dfrac{20}{10}=2\)

Theo tính chất

=> \(\dfrac{x}{3}=2\Rightarrow x=3.2\Rightarrow x=6\)

\(\dfrac{y}{7}=2\Rightarrow x=7.2\Rightarrow x=14\)

Vậy : x = 6 , y = 14

Answer 4

b)$\frac{x}{5}$ =$\frac{y}{2}$ và x-y=6

=> \(\dfrac{x-y}{5-2}=\dfrac{6}{3}=2\)

Theo tính chất

=> \(\dfrac{x}{5}=2\Rightarrow x=5.2\Rightarrow x=10\)

\(\dfrac{y}{2}=2\Rightarrow y=2.2\Rightarrow y=4\)

Vậy x= 10 , y =4

Answer 5

a) Đặt \(\dfrac{x}{3}=\dfrac{y}{7}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=7k\end{matrix}\right.\)

Ta có: 3k + 7k = 20

\(\Rightarrow\)10k = 20

\(\Rightarrow\) k = 2

Thay k = 2, ta có:

\(\left\{{}\begin{matrix}x=3.2\\y=7.2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=21\end{matrix}\right.\)

Vậy ...

b) Đặt \(\dfrac{x}{5}=\dfrac{y}{2}=k\)

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

Ta có: \(\text{5k - 2k = 6}\)

\(\Rightarrow3k=6\)

\(\Rightarrow k=2\)

Thay k = 2, ta có:

\(\left\{{}\begin{matrix}x=5.2\\y=2.2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=10\\y=4\end{matrix}\right.\)

Vậy ...