\(5x=7y\Rightarrow5.3y=7y\left(x=3y\right)\Rightarrow\)\(15y=7y\Rightarrow15y-7y=0\)\(\Rightarrow8y=0\Rightarrow y=0\Rightarrow x=0\)

 \(\frac{x}{2}\)= \(\frac{y}{3}\); \(\frac{y}{4}\)= \(\frac{z}{5}\)và x + y - z = 10

\(\Rightarrow\)\(\frac{x}{8}\)= \(\frac{y}{12}\); \(\frac{y}{12}\)= \(\frac{z}{15}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau: \(\frac{x}{8}\)= \(\frac{y}{12}\)= \(\frac{z}{15}\)= \(\frac{x+y-z}{8+12-15}\)= \(\frac{10}{5}\)= 2

\(\hept{\begin{cases}\frac{x}{8}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)

Vậy x= 16

       y= 24

       z= 30

d) 2x = 3y ; 5x = 7z và 3x - 7y + 5x = 3

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

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

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

Áp dụng tính chất dãy tỉ  số bằng nhau: \(\frac{x}{21}\)= \(\frac{y}{14}\)= \(\frac{z}{15}\)\(\Rightarrow\)\(\frac{3x}{63}\)= \(\frac{7y}{98}\)= \(\frac{5z}{75}\)= \(\frac{3x-7y+5z}{63-98+75}\)= \(\frac{30}{40}\)=\(\frac{3}{4}\)

\(\hept{\begin{cases}\frac{x}{21}=\frac{3}{4}\\\frac{y}{14}=\frac{3}{4}\\\frac{z}{15}=\frac{3}{4}\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=\frac{63}{4}\\y=\frac{21}{2}\\z=\frac{45}{4}\end{cases}}\)

Vậy x= \(\frac{63}{4}\)

      y= \(\frac{21}{2}\)

      z= \(\frac{45}{4}\)

c) \(4x=7y\Rightarrow\frac{x}{7}=\frac{y}{4}\Rightarrow\frac{x^2}{49}=\frac{y^2}{16}=\frac{x^2+y^2}{49+16}=\frac{260}{65}=4\)

\(\Rightarrow\orbr{\begin{cases}x^2=4.49=14^2\\y^2=4.16=8^2\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=14\\y=8\end{cases}}\)

d) \(\frac{x}{2}=\frac{y}{4}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}\Rightarrow\frac{x^2.y^2}{4.16}=\frac{x^4}{16}=\frac{4}{64}=\frac{1}{16}\Rightarrow x=1;y=2\)

a) Ta có:

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{-2}\) và \(5x-y+3z=-16\)

\(\Rightarrow\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}\)

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

\(\frac{5x}{15}=\frac{y}{5}=\frac{3z}{-6}=\frac{5x-y+3z}{15-5+\left(-6\right)}=\frac{-16}{4}=-4\)

\(\Rightarrow\frac{5x}{15}=-4\Rightarrow5x=\left(-4\right).15=-60\Rightarrow x=60:5=12\)

\(\Rightarrow\frac{y}{5}=-4\Rightarrow y=\left(-4\right).5=-20\)

\(\Rightarrow\frac{3z}{-6}=-4\Rightarrow3z=\left(-4\right).\left(-6\right)=24\Rightarrow y=24:3=8\)

Vậy ___________________________________________________________

b) Ta có:

\(2x=3y;5y=7z\)và \(3x-7y+5z=30\)

\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)\(=\frac{x}{21}=\frac{y}{14};\frac{y}{14}=\frac{z}{10}=\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

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

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

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

Từ đây dễ òi, bạn tự tìm x ; y ; z nhé :)

a) \(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{x+y+z}{10+6+21}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)

-> \(5x=2\cdot50=100\) => \(x=\frac{100}{5}=20\)

\(y=2\cdot6=12\)

\(2z=2\cdot42=84\) => \(z=\frac{84}{2}=42\)

\(\frac{1+3y}{12}=\frac{1+3y}{5x}=\frac{1+7y}{4x}\)(1)

Từ \(\frac{1+3y}{12}=\frac{1+3y}{5x}\)\(\Rightarrow5x=12\)\(\Rightarrow x=\frac{12}{5}\)

Thay \(x=\frac{12}{5}\)vào (1) ta có: \(\frac{1+3y}{12}=\frac{1+7y}{4.\frac{12}{5}}=\frac{1+7y}{\frac{48}{5}}=\frac{5\left(1+7y\right)}{48}\)

\(\Rightarrow\frac{4\left(1+3y\right)}{48}=\frac{5\left(1+7y\right)}{48}\)

\(\Rightarrow4\left(1+3y\right)=5\left(1+7y\right)\)\(\Leftrightarrow4+12y=5+35y\)

\(\Leftrightarrow23y=-1\)\(\Leftrightarrow y=-\frac{1}{23}\)

Vậy \(x=\frac{12}{5}\)và \(y=-\frac{1}{23}\)

...có đáp án mik đã đến muộn 6 năm

Ta có: 6x - 2y = 7y - 3x

=> 6x + 3x = 7y + 2y

=> 9x = 9y => x = y

=> x - y = 0

mà x - y = 10 (đb)

=> ko có x; t tm

7x - 2y = 5x - 3y

=> 7x - 5x = -3y + 2y

=> 2x = -y

=> \(\frac{x}{-1}=\frac{y}{2}\) => \(\frac{2x}{-2}=\frac{3y}{6}\)

áp dụng t/c của dãy tỉ số bằng nhau, ta có:

 \(\frac{2x}{-2}=\frac{3y}{6}=\frac{2x+3y}{-2+6}=\frac{20}{4}=5\)

=> \(\hept{\begin{cases}\frac{x}{-1}=5\\\frac{y}{2}=5\end{cases}}\) => \(\hept{\begin{cases}x=5.\left(-1\right)=-5\\y=5.2=10\end{cases}}\)

ta có 6x-2y=7y-3x chuyển vế sang

=>9x=9y

do x-y=10 nên x=10+y

=>9(10+y)=9y

=>90+9y=9y 

=>90=0y

=>y=0=>x=10

mik ko biét

Ta có:

\(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\) (1).

\(5x=7z\Rightarrow\frac{x}{7}=\frac{z}{5}\) (2).

Từ (1) và (2) \(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{x}{7}=\frac{z}{5}.\)

Có:

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

\(\frac{x}{7}=\frac{z}{5}\Rightarrow\frac{x}{21}=\frac{z}{15}.\)

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

\(\Rightarrow\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{75}\) và \(3x-7y+5z=30.\)

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

\(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{75}=\frac{3x-7y+5z}{63-98+75}=\frac{30}{40}=\frac{3}{4}.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{21}=\frac{3}{4}\Rightarrow x=\frac{3}{4}.21=\frac{63}{4}\left(KTM\right)\\\frac{y}{14}=\frac{3}{4}\Rightarrow y=\frac{3}{4}.14=\frac{21}{2}\left(KTM\right)\\\frac{z}{15}=\frac{3}{4}\Rightarrow z=\frac{3}{4}.15=\frac{45}{4}\left(KTM\right)\end{matrix}\right.\)

Vậy không có cặp số \(\left(x;y;z\right)\) nào thỏa mãn đề bài.

Chúc bạn học tốt!

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

\(m=\frac{5x+3y}{6x-7y}=\frac{10x+6y}{12x-14y}=\frac{10x+9x}{12x-21x}=\frac{19x}{-9x}=-\frac{19}{9}\)