Áp dụng tc của dãy tỉ số bằng nhau ta có :

\(\frac{x}{20}=\frac{y}{15}=\frac{y-x}{15-20}=\frac{20}{-5}=-4\)

\(\Rightarrow\begin{cases}x=-80\\y=-60\end{cases}\)

Theo t/c của dãy tỉ số bằng nhau ta có 

\(\frac{x}{20}=\frac{y}{15}=\frac{y-x}{15-20}=\frac{20}{-5}=-4\)

\(\frac{x}{20}=-4\Rightarrow x=-4.20=-80\)

\(\frac{y}{15}=-4\Rightarrow y=-4.15=-60\)

mk hoc lop 6 soryy ko giup duoc

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

ĐK: y\(\ne0;x\ne0,-15,3\)

\(\left\{{}\begin{matrix}\frac{y}{x}-\frac{y}{x+15}=\frac{1}{5}\\\frac{y}{x-3}-\frac{y}{x}=\frac{1}{20}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}y\left(\frac{1}{x}-\frac{1}{x+15}\right)=\frac{1}{5}\\y\left(\frac{1}{x-3}-\frac{1}{x}\right)=\frac{1}{20}\end{matrix}\right.\)\(\Leftrightarrow\frac{1}{5}:\left(\frac{1}{x}-\frac{1}{x+15}\right)=\frac{1}{20}:\left(\frac{1}{x-3}-\frac{1}{x}\right)\Leftrightarrow\frac{1}{5}:\frac{15}{x^2+15x}=\frac{1}{20}:\frac{3}{x^2-3x}\Leftrightarrow\frac{x^2+15x}{75}=\frac{x^2-3x}{60}\Leftrightarrow\frac{4x^2+60x}{300}=\frac{5x^2-15x}{300}\Leftrightarrow4x^2+60x=5x^2-15x\Leftrightarrow x^2-75x=0\Leftrightarrow x\left(x-75\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\left(ktm\right)\\x=75\left(tm\right)\end{matrix}\right.\)\(\Leftrightarrow\)\(y=\frac{1}{5}:\left(\frac{1}{75}-\frac{1}{75+15}\right)=90\)

Vậy (x;y)=(75;90)

=>\(\frac{x+y-z}{15+20-28}=\frac{7}{7}=1\)

=>\(\frac{x}{15}=1=>x=15\)

=>\(\frac{y}{20}=1=>y=20\)

=>\(\frac{z}{28}=1=>z=28\)

vậy:\(x=15;y=20;z=28\)

\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{x+y-z}{15+20-28}=\frac{7}{7}=1\)

\(\frac{x}{15}=1\Rightarrow x=1.15\Rightarrow x=15\)

\(\frac{y}{20}=1\Rightarrow y=1.20\Rightarrow y=20\)

\(\frac{z}{28}=1\Rightarrow z=1.28\Rightarrow z=28\)

Vậy x = 15

       y = 20

       z = 28 

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{15}=\frac{y}{20}=\frac{z}{28}=\frac{x+y-z}{15+20-28}=\frac{7}{7}=1\)

\(\Rightarrow\begin{cases}x=15\\y=20\\z=28\end{cases}\)

\(\frac{15}{x-9}=\frac{20}{y-12}\Leftrightarrow15\left(y-12\right)=20\left(x-9\right)\Leftrightarrow15y-180=20x-180\Leftrightarrow15y=20x\Leftrightarrow\frac{y}{20}=\frac{x}{15}\Leftrightarrow\frac{xy}{20}=\frac{x^2}{15}\Leftrightarrow x^2=\frac{15.1200}{20}=900\Leftrightarrow\orbr{\begin{cases}x=30\\x=-30\end{cases}}\)

Chia từng trường hợp tìm y, z.

\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{24}{z-24}\Rightarrow\)\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}=k\)

Do x . y = 1200 => ( 15k + 9) ( 20k + 12) = 1200

3(5k+3).4(5k+3) = 1200

12 ( 5k+3)² = 1200

( 5k+3)² =  100 hoặc ( 5k+3)² =  -100

=> 5k+3 = 10 hoặc 5k+3 = -10

=> 5k = 7 hoặc 5k = -13

=> k = 7/5 hoặc k = -13/5

 Vậy

\(\hept{\begin{cases}\text{x = 15 . \frac{7}{5}+9 = 30}\\y=20.\frac{7}{5}+12=40\\z=40.\frac{7}{5}+24=80\end{cases}}\)\(\hept{\begin{cases}x=15.\frac{7}{5}+9=30\\y=20.\frac{7}{5}+12=40\\z=40.\frac{7}{5}+24=80\end{cases}}\)hoặc\(\hept{\begin{cases}x=15.\frac{-13}{5}+9=-30\\y=20.\frac{-13}{5}+12=-40\\z=40.\frac{-13}{5}+24=-80\end{cases}}\)