Question

Tìm x, y biết :  3x=2y , 7y=5z và x-y+z=32

Answer 1

\(3x=2y;7y=5z\) va x-y+z=32

\(\Rightarrow3x=2y=\frac{x}{2}=\frac{y}{3}\)

\(\Rightarrow7y=5z=\frac{y}{5}=\frac{z}{7}\)

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

Suy ra  : \(\frac{x}{10}=2\Rightarrow x=2.10=20\)

\(\frac{y}{15}=2\Rightarrow y=2.15=30\)

\(\frac{z}{21}=3\Rightarrow z=3.21=63\)

Answer 2

3x=2y=>x/2=y/3=>x/10=y/15

7y=5z=>y/5=z/7=>y/15=z/21

=>x/10=y/15=z/21=x-y+z/10-15+21=32/16=2

=>x=20;y=30;z=42

vậy x=20;y=30;z=42

Answer 3

\(3x=2y;7y=5z\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{y}{15};\frac{y}{15}=\frac{z}{21}\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x-y+z}{10-15+21}=\frac{32}{16}=2\)

suy ra:

\(\frac{x}{10}=2\Rightarrow x=2.10=20\)

\(\frac{y}{15}=2\Rightarrow y=2.15=30\)

\(\frac{z}{21}=2\Rightarrow z=2.21=42\)

Answer 4

x=20

y=20

z=63

Answer 5

x=20

y= 30

z=42

Answer 6

Em Ko Biết😊

☺☺☺🙂☺☺☺☺☺☺☺☺🙂

   

Answer 7

thanks

Answer 8

20;30;42