2x=3y;5y=7z

=>x/3=y/2;y/7=z/5

=>x/21=x/14;y/14=z/10

=>x/21=y/14=z/10

=>3x/63=7y/98=5z/50

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

3x/63=7y/98=5z/50=3x-7y+5z/63-98+50=30/15=2

suy ra : 3x/63=2 =>3x=126 =>x=126:3=42

7y/98=2 =>7y =196 =>y=196:7=28

5z/50=2 =>5z = 100 => z=100:5=20

\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\left(1\right)\)

\(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\left(2\right)\)

Từ 1 và 2 

=> \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Áp dụng tính chất dãy tỉ số = nhau ta có

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2\)

\(\frac{x}{21}=2\Rightarrow x=42\)

\(\frac{y}{14}=2\Rightarrow y=28\)

\(\frac{z}{10}=2\Rightarrow z=20\)

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

\(\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{3}=\frac{7y}{14};\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{2y}{14}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\Rightarrow\frac{3x}{63}=\frac{5y}{70}=\frac{7z}{70}\)

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

\(\frac{3x}{63}=\frac{5y}{70}=\frac{7z}{70}=\frac{3x+5y-7z}{63+70-70}=\frac{30}{63}=\frac{10}{21}\)

\(\frac{3x}{63}=\frac{10}{21}\Rightarrow x=\frac{10}{21}.63:3=10\)

\(\frac{5y}{70}=\frac{10}{21}\Rightarrow y=\frac{10}{21}.70:5=\frac{20}{3}\)

\(\frac{7z}{70}=\frac{10}{21}\Rightarrow z=\frac{10}{21}.70:7=\frac{100}{21}\)

1,7y

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

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

\(\hept{\begin{cases}\frac{x}{2}=\frac{x}{3}\\\frac{y}{5}=\frac{x}{7}\end{cases}\Rightarrow}\frac{x}{2}=\frac{5y}{15};\frac{3y}{15}=\frac{z}{7}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{21}\)

Áp dụng tính chát dãy tỉ số = nhau ta có:

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

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

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

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

b, Tự làm

c, \(5x=2y\Leftrightarrow\frac{x}{2}=\frac{y}{5}\)

\(2x=3z\Leftrightarrow\frac{x}{3}=\frac{z}{2}\)

\(\Leftrightarrow\frac{x}{2}=\frac{y}{5};\frac{x}{3}=\frac{z}{2}\)

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{x}{6}=\frac{z}{10}\)

\(\Leftrightarrow\frac{x}{6}=\frac{y}{15}=\frac{z}{10}\)

Đặt \(\frac{x}{6}=\frac{y}{15}=\frac{z}{10}=k(k\inℤ)\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\)

\(\Leftrightarrow x\cdot y=6k\cdot15k=90\)

\(\Leftrightarrow90:k^2=90\Leftrightarrow k^2=1\Leftrightarrow k=\pm1\)

\(\Leftrightarrow\hept{\begin{cases}x=6k\\y=15k\\z=10k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=15\\z=10\end{cases}}\)hay \(\hept{\begin{cases}x=-6\\y=-15\\z=-10\end{cases}}\)

Vậy \((x,y)\in(6,15);(-6,-15)\)

d, \(2x=3y=5z\Leftrightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\)

\(\Leftrightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}\)

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

\(\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

Vậy : \(\hept{\begin{cases}\frac{x}{15}=5\\\frac{y}{10}=5\\\frac{z}{6}=5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=75\\y=50\\z=30\end{cases}}\)

Lời giải:

$2x=3y\Rightarrow \frac{x}{3}=\frac{y}{2}$

$\Rightarrow \frac{x}{21}=\frac{y}{14}$

$5y=7z\Rightarrow \frac{y}{7}=\frac{z}{5}\Rightarrow \frac{y}{14}=\frac{z}{10}$

Vậy:

$\frac{x}{21}=\frac{y}{14}=\frac{z}{10}$

$=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=\frac{30}{15}=2$

$\Rightarrow x=21.2=42; y=14.2=28; z=10.2=20$

2x = 3y => 10x=15y
5y = 7z => 15y=21z
=> 10x=15y=21z =>x=2,1z
y=1,4z
Mà : 3x - 7y + 5z = 30 => 6,3z - 9,8z + 5z=30 =>1,5z=30
=>z=20
y=28
x=42

Từ \(2x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{x}{3}.\frac{1}{7}=\frac{y}{2}.\frac{1}{7}=\frac{x}{21}=\frac{y}{14}\)( 1 )

Từ \(5y=7z\)\(\Rightarrow\)\(\frac{y}{7}=\frac{z}{5}=\frac{y}{7}.\frac{1}{2}=\frac{z}{5}.\frac{1}{2}=\frac{y}{14}=\frac{z}{10}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\)\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Đặt \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=k\)

\(\Rightarrow\hept{\begin{cases}x=21k\\y=14k\\z=10k\end{cases}}\)

Thay vào \(3x+5z-7y=30\)ta có ;

\(3.21k+5.10k-7.14k=30\)

\(63k+50k-98k=30\)

\(15k=30\)

\(k=2\)

Thay vào ta được :

\(\Rightarrow\hept{\begin{cases}x=21.2\\y=14.2\\z=10.2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=42\\y=28\\z=20\end{cases}}\)