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:

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

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

Vi 2x+7y-3z=2 nen 4x+14y-6z=4

                           3y+14y-3y=4

                           14y=4 nen y=2/7

                                           x=3/14

                                           z=1/7

                           Bn thu tinh lai gium mk nhe !

\(\dfrac{x+2y}{4x-3y}=-2\)

=>x+2y=-8x+6y

=>9x=4y

hay x/y=4/9

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

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{98}{48}=\frac{49}{23}\)

suy ra :

\(\frac{x}{10}=\frac{49}{23}\Rightarrow x=\frac{490}{23}\)

\(\frac{y}{15}=\frac{49}{23}\Rightarrow y=\frac{735}{23}\)

\(\frac{z}{21}=\frac{49}{23}\Rightarrow z=\frac{1029}{23}\)

bạn xem lại đề ra số hơi xấu

oanh ngố dễ vậy mà ko làm được

đã học đâu mà biết hôm nay mới học nè ! 

a, x/5-y/2

=> 3x/15=2y/4=3x-2y/15-4=44/11=4

+, x/5=4 => x=20

+, y/2=4 => y=8

c, 4x=3y

=> x/3=y/4=x-y=3-4=11/-1=-11

+, x/3=-11 => x=-33

+, y/4=-11 => y=-44

4x = 3y nên \(\frac{x}{3}\)=\(\frac{y}{4}\)

Theo tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{3}\)=\(\frac{y}{4}\)=\(\frac{x-y}{3-4}\)=\(\frac{11}{-1}\)= -11

+) \(\frac{x}{3}\)= -11

      x = -11x3=-33

+) \(\frac{y}{4}\)= -11

      y = -11x4 =-44

 Vậy x = -33 ; y = -44

\(\hept{\begin{cases}4x-3z=z\\6y-x=z\\2x+3y+4z=19\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=z\\6y-x=z\\2x+3y+4z=19\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=z\\3y=z=x\\2x+3y+4z=19\end{cases}}\)

\(\Leftrightarrow\)2x+x+4x=19 \(\Leftrightarrow\)x=z = \(\frac{19}{7}\)

y=\(\frac{19}{21}\)

x= \(\frac{19}{7}\)

y= \(\frac{19}{21}\)

z= \(\frac{19}{7}\)

bn làm rõ dc ko