\(x:y=19:21\Leftrightarrow\dfrac{x}{y}=\dfrac{19}{21}\Leftrightarrow\dfrac{x}{21}=\dfrac{y}{19}\)

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

\(\dfrac{x}{21}=\dfrac{y}{19}=\dfrac{x-y}{21-19}=\dfrac{4}{2}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{21}=2\\\dfrac{y}{19}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=42\\y=38\end{matrix}\right.\)

Vậy ...

b/ \(3x=5y=7z\)

\(\Leftrightarrow\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}\)

\(\Leftrightarrow\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

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

\(\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y+z}{35+21+15}=\dfrac{41}{71}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{35}=\dfrac{41}{71}\\\dfrac{y}{21}=\dfrac{41}{71}\\\dfrac{z}{15}=\dfrac{41}{71}\end{matrix}\right.\) (tự tính yieeps)

Vậy ..

có 3x=5y=7z

\(\Rightarrow\frac{x}{35}=\frac{y}{21}=\frac{x}{15}\) (z/15 nha, ko phải x/15)

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

\(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}=\frac{x+y-z}{35+21-15}=\frac{41}{41}=1\)

=>\(\frac{x}{35}=1\Rightarrow x=35\)

\(\frac{y}{21}=1\Rightarrow y=21\)

\(\frac{z}{15}=1\Rightarrow z=15\)

vậy...........

x=35;y=21;z=15

từ 3x=5y=7z=x/3=y/5=z/7

=>\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=\frac{x+y-z}{3+5-7}=\frac{41}{1}=41\)

\(\Rightarrow x=3x41=123\)

\(y=41x5=205\)

\(z=41x7=287\)

tự vậy nhé

Theo đề bài ta có:

\(3x=5y=7z\Leftrightarrow3x.\dfrac{1}{105}=5y.\dfrac{1}{105}=7z.\dfrac{1}{105}\)

Hay \(\dfrac{3x}{105}=\dfrac{5y}{105}=\dfrac{7z}{105}\Leftrightarrow\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)

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

\(\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}=\dfrac{x+y-z}{35+21-15}=\dfrac{41}{41}=1\)

Nên \(\left\{{}\begin{matrix}x=1.35=35\\y=1.21=21\\z=1.15=15\end{matrix}\right.\)

Áp dụng TCCDTSBN, Ta có:

\(\dfrac{3x}{3.5.7}=\dfrac{5y}{3.5.7}=\dfrac{7z}{3.5.7}\)

\(\Rightarrow\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\)=\(\dfrac{x+y-z}{35+21-15}\)=\(\dfrac{41}{41}=1\)

\(\Rightarrow\dfrac{x}{35}=1\Rightarrow x=1.35=35\)

\(\dfrac{y}{21}=1\Rightarrow y=1.21=21\)

\(\dfrac{z}{15}=1\Rightarrow z=1.15=15\)

\(\Rightarrow\)x= 35; y= 21; z=15

Ta có: \(3x=5y=7z\) \(\Leftrightarrow\) \(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{7}}\) và \(x+y-z=41\)

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

\(\dfrac{x}{\dfrac{1}{3}}=\dfrac{y}{\dfrac{1}{5}}=\dfrac{z}{\dfrac{1}{7}}=\dfrac{x+y-z}{\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}}=\dfrac{41}{\dfrac{41}{105}}=105\)

\(\dfrac{x}{\dfrac{1}{3}}=105\Rightarrow x=105.\dfrac{1}{3}=35\)

\(\dfrac{y}{\dfrac{1}{5}}=105\Rightarrow y=105.\dfrac{1}{5}=21\)

\(\dfrac{z}{\dfrac{1}{7}}=105\Rightarrow z=105.\dfrac{1}{7}=15\)

Vậy \(x=35\); \(y=21\); \(z=15\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\) , \(\dfrac{y}{7}=\dfrac{z}{5}\) \(\Rightarrow\dfrac{x}{35}=\dfrac{y}{21},\dfrac{y}{21}=\dfrac{z}{15}\) \(\Rightarrow\dfrac{x}{35}=\dfrac{y}{21}=\dfrac{z}{15}\) \(=\) \(\dfrac{x+y-z}{35+21-15}\) = \(\dfrac{41}{11}\) ta có \(\dfrac{x}{35}=\dfrac{41}{11}\Rightarrow x=41\times35\div11=130,\left(45\right)\) \(y=130,\left(45\right)\times3\div5\) \(=78,\left(27\right)\) \(z=78.\left(27\right)\times5\div7=55.\left(90\right)\)

suy ra:x/2=y/3  ,  y/7 = z/5

suy ra x/14 = y/21 = z/15 = 3x/42 = 5y/105  = 7z/105

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

3x/42= 5y/105 = 7z/105= 3x +5y -7z/42+105-105 = 10/7

suy ra : x= 20

y = 30

z = 150/7

Néu đúng thì k cho mk nha

ta có: 3x=2y => \(\frac{x}{2}=\frac{y}{3}\)

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

=>\(\frac{x}{14}=\frac{y}{21}=\frac{z}{15}\)

=>\(\frac{3x}{42}=\frac{5y}{105}=\frac{7z}{245}=\)\(\frac{3x+5y-7z}{42+105-105}=\frac{60}{42}=\frac{10}{7}\)

\(\frac{x}{14}=\frac{10}{7}\)=> x =20

\(\frac{y}{21}=\frac{10}{7}\)=> y = 30

\(\frac{z}{15}=\frac{10}{7}\) => z=\(\frac{150}{7}\)

Đáp số:20;30;\(\frac{150}{7}\)

\(3x=5y=7z\)

=>  \(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}\)

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

\(\frac{x}{35}=\frac{y}{21}=\frac{z}{15}=\frac{x+y-z}{35+21-15}=\frac{41}{41}=1\)

đến đây tự tính nhé

Đặt : \(3x=5y=7x=k\)

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

\(\Rightarrow x+y-z=\frac{k}{3}+\frac{k}{5}-\frac{k}{7}=\frac{41k}{105}=41\)

\(\Rightarrow41k=41.105\)

\(\Rightarrow k=105\)

\(\Rightarrow x=\frac{105}{3}=35;\)\(y=\frac{105}{5}=21;\)\(z=\frac{105}{7}=15\)

----------------------HOK TỐT---------------------------

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

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

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

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

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

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

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

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