3) \(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.9=90\end{matrix}\right.\)

4) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x}{10}=\dfrac{2y}{6}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=7.3=21\end{matrix}\right.\)

5) \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-z}{5+7-10}=\dfrac{20}{2}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.5=50\\y=10.7=70\\z=10.10=100\end{matrix}\right.\)

6) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2y}{8}=\dfrac{2z}{10}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\Rightarrow\left\{{}\begin{matrix}x=11.3=33\\y=11.4=44\\z=11.5=55\end{matrix}\right.\)

7) \(\Rightarrow\dfrac{x}{12}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{12+6-10}=\dfrac{20}{8}=\dfrac{5}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}.12=30\\y=\dfrac{5}{2}.6=15\\z=\dfrac{5}{2}.10=25\end{matrix}\right.\)

Câu 3:

\(\dfrac{x}{y}=\dfrac{5}{9}\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)

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

\(\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{x-y}{5-9}=\dfrac{-40}{-4}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=5\\ \dfrac{y}{9}=10\Rightarrow y=90\)

Câu b:

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{5x-2y}{10-6}=\dfrac{28}{4}=7\)

\(\dfrac{x}{2}=7\Rightarrow x=14\\ \dfrac{y}{3}=7\Rightarrow y=21\)

Câu c:

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

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{10}=\dfrac{x+y-1}{5+7-10}=\dfrac{20}{2}=10\)

\(\dfrac{x}{5}=10\Rightarrow x=50\\ \dfrac{y}{7}=10\Rightarrow y=70\\ \dfrac{z}{10}=10\Rightarrow z=100\)

Câu d:

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

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{3x-2y+2z}{9-8+10}=\dfrac{121}{11}=11\)

\(\dfrac{x}{3}=11\Rightarrow x=3\\ \dfrac{y}{4}=11\Rightarrow y=44\\ \dfrac{z}{5}=11\Rightarrow z=55\)

Câu e:

\(\dfrac{x}{4}=\dfrac{y}{2}\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}\\\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{y}{6}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10} \)

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

\(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{x+y-z}{8+6-10}=\dfrac{20}{4}=5\)

\(\dfrac{x}{8}=5\Rightarrow x=40\\ \dfrac{y}{6}=5\Rightarrow y=30\\ \dfrac{z}{10}=5\Rightarrow z=50\)

x/3=y/5=x+y/3+5=16/8=2

x/3=2 suy ra x=6

y/5=2 suy ra y=10

x/2=y/3suy ra x/8=y/12

y/4=z/5 suy ra y/12=z/15

x/8=y/12=z/15=x+y-z/8+12-15=10/5=2

x/8=2 suy ra x=16

y/12=2 suy ra y=24

x/15=2 suy ra z=30

\(\frac{x}{2}=\frac{y}{3}\Leftrightarrow\frac{x}{8}=\frac{y}{12};\frac{y}{4}=\frac{z}{5}\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)

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

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

x/8 = 2 => x = 16

y/12 = 2 => y = 24

z/15 = 2 => z = 30

Vậy x = 16 ; y = 24 ; z = 30

áp dụng tính chất dãy tỉ số bằng nhau
 

Ta có: x/2 = y/3 ; y/4 = z/5 và x+y-z=10

=> x/8=y/12=z/15

ADTC dãy tỷ số = nhau ta có:

x/8=y/12=z/15=(x+y-z)/(8+12-15)=10/5=2

Suy ra: x/8=2=>x=16

y/12=2=>y=24

z/15=2=>z=30

Vậy....

HT

\(\frac{x}{2}=\frac{y}{3};\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x}{8}=\frac{y}{12};\frac{y}{12}=\frac{z}{15}\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

suy ra: \(\frac{x}{8}=2\Rightarrow x=2.8=16\)

\(\frac{y}{12}=2\Rightarrow y=2.12=24\)

\(\frac{z}{15}=2\Rightarrow z=2.15=30\)

Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{8}=\frac{y}{12}\)

\(\frac{y}{4}=\frac{z}{5}\) => \(\frac{y}{12}=\frac{z}{15}\)

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

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

\(\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)

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

\(\frac{x}{2}=\frac{y}{3}\)     \(\left(\text{*}\right)\)

\(\frac{y}{4}=\frac{z}{5}\)       \(\left(\text{*}\text{*}\right)\)

\(x+y-z=10\)     \(\left(\text{*}\text{*}\text{*}\right)\)

\(\left(\text{*}\right)\)\(\Leftrightarrow3x=2y\Leftrightarrow x=\frac{2y}{3}\)

\(\left(\text{*}\text{*}\right)\)\(\Leftrightarrow5y=4z\Leftrightarrow z=\frac{5y}{4}\)  

Cả (*) và (**) thế vào (***)

\(\frac{2y}{3}+y-\frac{5y}{4}=10\Leftrightarrow\frac{5y}{12}=10\Leftrightarrow y=24\)

\(\Leftrightarrow x=16;z=30\)

Vậy ...