\(\frac{4}{x-3}\)=\(\frac{8}{y-6}\)=\(\frac{20}{z-15}\)

=> \(\frac{x-3}{4}\)=\(\frac{y-6}{8}\)=\(\frac{z-15}{20}\)

=> \(\frac{x}{4}\)-\(\frac{3}{4}\)= \(\frac{y}{8}\)-\(\frac{6}{8}\)=\(\frac{z}{20}\)-\(\frac{15}{20}\)

=> \(\frac{x}{4}\)=\(\frac{y}{8}\)=\(\frac{z}{20}\)

Đặt \(\frac{x}{4}\)=\(\frac{y}{8}\)=\(\frac{z}{20}\)=k

\(\frac{x}{4}\)= k => x = 4 . k

\(\frac{y}{8}\)= k => y = 8 . k

\(\frac{z}{20}\)= k => z = 20 . k

Mà x.y.x = 640

(4k) . (8k) . (20k)= 640

640 . kmũ3 = 640

k mũ 3 = 640:640

k mũ 3 = 1

\(\frac{x}{4}\)= 1 => x = 4 . 1 = 4

\(\frac{y}{8}\)= 1 => y = 8 . 1 = 8

\(\frac{z}{20}\)= 1 => z = 20 . 1=20

Vậy x=4, y=8, z=20

k mình nha,đúng 100%

Ta có:\(\frac{4}{x-3}=\frac{8}{y-6}=\frac{20}{z-15}\)

=>\(\frac{x-3}{4}=\frac{y-6}{8}=\frac{z-15}{20}\)

=>\(\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{6}{8}=\frac{z}{20}-\frac{15}{20}\)

=>\(\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{3}{4}=\frac{z}{20}-\frac{3}{4}\)

=>\(\frac{x}{4}=\frac{y}{8}=\frac{z}{20}\)

Đặt \(\frac{x}{4}=\frac{y}{8}=\frac{z}{20}=k\Rightarrow x=4k,y=8k,z=20k\)

Thay vào đề ta có: xyz = 640

=> 4k.8k.20k = 640

=> 640k³ = 640

=> k³ = 1

=> k = 1

=> x = 4, y = 8, z = 20

Vậy...

ai làm cho mik đi

\(a)\)Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{2x+3}{5x+2}=\frac{4x+5}{10x+2}=\frac{2\cdot(2x+3)-(4x+5)}{2\cdot(5x+2)-(10x+2)}=\frac{4x+6-4x-5}{10x+4-10x-2}=\frac{1}{2}\)

Suy ra :

\(\frac{2x+3}{5x+2}=\frac{1}{2}\Rightarrow1\cdot(5x+2)=2\cdot(2x+3)\)

\(5x+2=4x+6\)

\(5x-4x=6-2\)

\(x=4\)

\(b)\)Ta có : \(\frac{4}{x-3}=\frac{8}{y-6}=\frac{20}{z-15}\)

\(\Rightarrow\frac{x-3}{4}=\frac{y-6}{8}=\frac{z-15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{6}{8}=\frac{z}{20}-\frac{15}{20}\)

\(\Rightarrow\frac{x}{4}-\frac{3}{4}=\frac{y}{8}-\frac{3}{4}=\frac{z}{20}-\frac{3}{4}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{8}=\frac{z}{20}\)

Đặt : \(\frac{x}{4}=\frac{y}{8}=\frac{z}{20}=k\Rightarrow x=4k;y=8k;z=20k\)

Thay vào đề , ta có : xyz = 640

\(\Rightarrow4k\cdot8k\cdot20k=640\)

\(\Rightarrow640k^3=640\)

\(\Rightarrow k^3=1\)

\(\Rightarrow k=1\)

\(\Rightarrow x=4;y=8;z=20\)

Vậy

cam on nha

áp dụng DSTCBN:

Ta có:

\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

\(\Rightarrow\frac{x-30}{10}=\frac{y-15}{5}=\frac{z-21}{7}\)

\(\frac{\Rightarrow x}{10}-\frac{30}{10}=\frac{y}{5}-\frac{15}{5}=\frac{z}{7}-\frac{21}{7}\)

\(\frac{\Rightarrow x}{10}-3=\frac{y}{3}-3=\frac{z}{7}-3\)

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

\(\frac{x}{10}=\frac{y}{5}=\frac{z}{7}=t=\hept{\begin{cases}x=10t\\y=5t\\z=7t\end{cases}}\)

\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)

\(\Rightarrow\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

\(\text{Ta có:}\)\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

            \(\Leftrightarrow\frac{x-30}{40}=\frac{y-15}{40}=\frac{z-21}{28}\)

             \(\Leftrightarrow\frac{x}{40}-\frac{30}{40}=\frac{y}{40}-\frac{15}{40}=\frac{z}{28}-\frac{21}{28}\)

              \(\Leftrightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)\

                \(\Leftrightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

          \(\text{đặt:}\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\)

\(\Rightarrow x=40k\)

\(\Rightarrow y=20k\)

\(\Rightarrow z=28k\)

\(\text{Theo đề ta có :}\)\(x.y.z=22400\Rightarrow40k.20k.28k=22400\)

                                 \(\Rightarrow22400.k^3=22400\)

                                 \(\Rightarrow k^3=1\)

                                  \(\Rightarrow k=\pm1\)

\(\text{Với k=1 thì :}\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

\(\text{Với k=-1 thì :}\)\(\hept{\begin{cases}x=-40\\y=-20\\z=-28\end{cases}}\)

Thiên tài thật: \(k^3=1\Rightarrow k=\pm1\)

Dẫn đến: \(\left(-40\right).\left(-20\right).\left(-28\right)=22400\)?????

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\Rightarrow\frac{6x}{11.18}=\frac{9y}{2.18}=\frac{18z}{5.18}\)

\(\Rightarrow\frac{-x}{-33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+y+z}{-33+4+5}=\frac{-120}{-24}=5\)

\(\Rightarrow x=165;y=20;z=25\)

a) Ta có:

\(\begin{array}{l}\frac{x}{3} = \frac{y}{4} \Rightarrow \frac{x}{3}.\frac{1}{5} = \frac{y}{4}.\frac{1}{5} \Rightarrow \frac{x}{{15}} = \frac{y}{{20}};\\\frac{y}{5} = \frac{z}{6} \Rightarrow \frac{y}{5}.\frac{1}{4} = \frac{z}{6}.\frac{1}{4} \Rightarrow \frac{y}{{20}} = \frac{z}{{24}}\end{array}\)

Vậy  \(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}}\) (đpcm)

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

\(\frac{x}{{15}} = \frac{y}{{20}} = \frac{z}{{24}} = \frac{{x - y + z}}{{15 - 20 + 24}} = \frac{{ - 76}}{{19}} =  - 4\)

Vậy x = 15 . (-4) = -60; y = 20. (-4) = -80; z = 24 . (-4) = -96

1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)

\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)

=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)

=>\(x=3\cdot20=60\)

    \(y=3\cdot24=72\)

    \(z=3\cdot21=63\)

3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)

=> \(x=1\cdot15=15\)

     \(y=1\cdot7=7\)

     \(z=1\cdot3=3\)

     \(t=1\cdot1=1\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

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

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\Rightarrow x=27;y=36;z=60\)

b, \(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

\(\Rightarrow x=18;y=24;z=30\)

c, \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-4}{4}=\frac{2x+3y-z-2-6+4}{4+9-4}=\frac{46}{9}\)

\(\Rightarrow x=\frac{101}{9};y=\frac{52}{3};z=\frac{220}{9}\)

d, Đặt \(x=2k;y=3k;z=5k\Rightarrow xyz=810\Rightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\Leftrightarrow k=3\)Với k = 3 thì \(x=6;y=9;z=15\)