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 \(\dfrac{x}{35}=\dfrac{41}{11}\Rightarrow x=41\times35\div11=130,\left(45\right)\) \(y=130,\left(45\right)\times3\div5\) \(=78,\left(27\right)\) \(z=78.\left(27\right)\times5\div7=55.\left(90\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

Giúp mik vs, cảm ơn mọi người

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

dãy số tỉ bằng nhau mà làm

Câu cuối đề chưa rõ ràng , mà cho dù có rõ cùng nên sử dụng đặt bằng k  

1) \(\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x-y+z}{8-12+15}=\dfrac{10}{11}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{10}{11}\\\dfrac{y}{12}=\dfrac{10}{11}\\\dfrac{z}{15}=\dfrac{10}{11}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{80}{11}\\y=\dfrac{120}{11}\\z=\dfrac{150}{11}\end{matrix}\right.\)

2) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{5}=\dfrac{z}{7}\end{matrix}\right.\) \(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{136}{62}=\dfrac{68}{31}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=\dfrac{68}{31}\\\dfrac{y}{20}=\dfrac{68}{31}\\\dfrac{z}{28}=\dfrac{68}{31}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1020}{31}\\y=\dfrac{1360}{31}\\z=\dfrac{1904}{31}\end{matrix}\right.\)

3) \(\Rightarrow\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}\)

Áp dụng t/c dtsbn:

\(\dfrac{3x-9}{15}=\dfrac{5y-25}{5}=\dfrac{7z+21}{49}=\dfrac{3x+5y-7z-9-25-21}{15+5-49}=-\dfrac{45}{29}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3x-9}{15}=-\dfrac{45}{29}\\\dfrac{5y-25}{5}=-\dfrac{45}{29}\\\dfrac{7z+21}{49}=-\dfrac{45}{29}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{138}{29}\\y=\dfrac{100}{29}\\z=-\dfrac{402}{29}\end{matrix}\right.\)