1: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{0,3}=\dfrac{y}{0.2}=\dfrac{z}{0.1}=\dfrac{x-y}{0.3-0.2}=\dfrac{1}{0.1}=10\)

Do đó: x=3; y=2; z=1

Áp dụng t/c dtsbn:

\(\dfrac{x}{12}=\dfrac{y}{13}=\dfrac{z}{15}=\dfrac{x+y+z}{12+13+15}=\dfrac{160}{40}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.12=48\\y=4.13=52\\z=4.15=60\end{matrix}\right.\)

1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)

⇒ x=4;y=6;z=8

\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)

\(2,\) Áp dụng t/c dtsbn

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)

\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

Áp dụng t/c dtsbn

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)

\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)

1.Tìm x, y.

2.TÌM x, y, z.

3.TÌM x, y, z.

1) \(\dfrac{x}{3}=\dfrac{y}{4}=k\)\(\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\end{matrix}\right.\)

\(\Rightarrow xy=12k^2=192\Rightarrow k=\pm4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm12\\y=\pm16\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=12\\y=16\end{matrix}\right.\\\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\end{matrix}\right.\)

2) Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{-90}{9}=-10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-10\right).2=-20\\y=\left(-10\right).3=-30\\z=\left(-10\right).5=-50\end{matrix}\right.\)

3) Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{8}=\dfrac{z}{5}=\dfrac{3x}{9}=\dfrac{2z}{10}=\dfrac{3x+y-2z}{9+8-10}=\dfrac{14}{7}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.3=6\\y=2.8=16\\z=2.5=10\end{matrix}\right.\)

Đặt \(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{40}=k\Leftrightarrow x=15k;y=20k;z=40k\)

\(xy=1200\\ \Leftrightarrow300k^2=1200\\ \Leftrightarrow k^2=4\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30;y=40;z=80\\x=-30;y=-40;z=-80\end{matrix}\right.\)

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

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k² = 90 => k² = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)