\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

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

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)

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

\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)

⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)

\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)

\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)

\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)

4x=3y, 5y=3z=>\(\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{12}=\frac{z}{20}\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

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

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

suy ra:

\(\frac{x}{9}=3\Rightarrow x=27\)

\(\frac{y}{12}=3\Rightarrow y=36\)

\(\frac{z}{20}=3\Rightarrow z=60\)

4x = 3y => \(\frac{x}{3}=\frac{y}{4}\) => \(\frac{x}{9}=\frac{y}{12}\) (1)

5y = 3z => \(\frac{y}{3}=\frac{z}{5}\) => \(\frac{y}{12}=\frac{z}{20}\)  (2)

(1);(2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{2.9-3.12+20}=\frac{6}{2}=3\) 

=> x = 3.9 = 27; b = 3.12 = 36; c = 3.20 = 60

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

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

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

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

=> x = (-2).21 = -42

     y = (-2).14 = -28

     z = (-2).10 = -20

Vậy ...

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

\(5y=7z\) \(\Rightarrow\)\(\frac{y}{7}=\frac{z}{5}\)hay  \(\frac{y}{14}=\frac{z}{10}\)

suy ra:   \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\) hay   \(\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}\)

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

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

suy ra:   \(\frac{3x}{63}=-2\)\(\Rightarrow\)\(x=-42\)

             \(\frac{7y}{98}=-2\)\(\Rightarrow\)\(y=-28\)

             \(\frac{5z}{50}=-2\) \(\Rightarrow\)\(z=-10\)

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:

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

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

2x + \(\frac{1}{7}\) = \(\frac{1}{y}\)

<=> \(\frac{1}{y}\) - 2x = \(\frac{1}{7}\)

<=> \(\frac{1}{y}\) - \(\frac{2xy}{y}\) = \(\frac{1}{7}\)

<=>  \(\frac{1-2xy}{y}\) = \(\frac{1}{7}\)

<=> 7(1-2xy) = y

<=> 7 -14xy  =y

<=> y+14xy   = 7

<=> y(14x+1) =7

vì x,y thuộc Z

nên y(14x+1) = 1.7=7.1=(-1)(-7)=(-7)(-1)

sau đó lập bảng nha bn

1.

\(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}\)

\(\frac{y}{5}=\frac{z}{7}=>\frac{y}{15}=\frac{z}{21}\)

=>\(\frac{x}{10}=\frac{y}{15}=\frac{z}{21}=\frac{x+y+z}{10+15+21}=\frac{92}{46}=2\)

=> x=2x10=20

y=2x15=30

z=2x21=42

2.

\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\frac{4x-3y-2z}{4-6-6}=\frac{36}{-8}=-\frac{9}{2}\)

=> x=\(-\frac{9}{2}x1=-\frac{9}{2}\)

y=\(-\frac{9}{2}x2=-9\)

z=\(-\frac{9}{2}x3=-\frac{27}{2}\)

\(4x=3y;5y=3z\Rightarrow\frac{x}{3}=\frac{y}{4};\frac{y}{3}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{12}=\frac{z}{20}\)

\(\Rightarrow\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\)

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

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

suy ra :

\(\frac{x}{9}=3\Rightarrow x=27\)

\(\frac{y}{12}=3\Rightarrow y=36\)

\(\frac{z}{20}=3\Rightarrow z=60\)

4x = 3y      => x/3 = y/4                 (1)

5y = 3z      => y/3 = z/5                  (2)

từ (1), (2)      => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}\) và 2x - 3y + z = 6

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

 \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{9\cdot2-3\cdot12+20}=\frac{6}{2}=3\)

suy ra: \(\frac{x}{9}=3\Rightarrow x=9\cdot3=27\)

\(\frac{y}{12}=3\Rightarrow y=12\cdot3=36\)

\(\frac{z}{20}=3\Rightarrow z=20\cdot3=60\)