b) Ta có: 7x=10y=12z

nên \(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}\)

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

\(\dfrac{x}{\dfrac{1}{7}}=\dfrac{y}{\dfrac{1}{10}}=\dfrac{z}{\dfrac{1}{12}}=\dfrac{x+y+z}{\dfrac{1}{7}+\dfrac{1}{10}+\dfrac{1}{12}}=\dfrac{685}{\dfrac{137}{420}}=2100\)

Do đó:

\(\left\{{}\begin{matrix}x=2100\cdot\dfrac{1}{2}=1050\\y=2100\cdot\dfrac{1}{10}=210\\z=2100\cdot\dfrac{1}{12}=175\end{matrix}\right.\)

a) ta có: \(5x=10y\Rightarrow\frac{x}{10}=\frac{y}{5}\Rightarrow\frac{x}{120}=\frac{y}{60}\)

\(10y=12z\Rightarrow\frac{y}{12}=\frac{z}{10}\Rightarrow\frac{y}{60}=\frac{z}{50}\)

\(\Leftrightarrow\frac{x}{120}=\frac{y}{60}=\frac{z}{50}\)

ADTCDTSBN

có: \(\frac{x}{120}=\frac{y}{60}=\frac{z}{50}=\frac{x-y+z}{120-60+50}=\frac{100}{110}=\frac{10}{11}\)

=> x/120 = 10/11 => ...

...

rùi bn tự lm típ nha!

b) ta có: \(\frac{x}{5}=\frac{y}{6}=\frac{z}{9}=k\Rightarrow\hept{\begin{cases}x=5k\\y=6k\\z=9k\end{cases}}\)

mà xyz = 270

=> 5k.6k.9k = 270

270.k³ = 270

=> k³ = 1

=> k = 1

=> x = 5k => x = 5

y = 6k => y = 6

z = 9k => z = 9

KL:...

\(\frac{3x}{5}=\frac{-5y}{4}=\frac{4z}{3}=\frac{6x}{10}=\frac{-10y}{8}=\frac{12z}{9}\)

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

\(\frac{6x+\left(-10y\right)+12z}{10+8+9}=\frac{30}{27}=\frac{10}{9}\)

suy ra \(\frac{3x}{5}=\frac{10}{9}\Rightarrow3x=\frac{50}{9}\Rightarrow x=\frac{50}{27}\)

\(\frac{-5y}{4}=\frac{10}{9}\Rightarrow-5y=\frac{40}{9}\Rightarrow y=\frac{-8}{9}\)

\(\frac{4z}{3}=\frac{10}{9}\Rightarrow4z=\frac{10}{3}\Rightarrow z=\frac{5}{6}\)

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

 \(\frac{3x\cdot2}{5\cdot2}=\frac{-5y\cdot\left(-2\right)}{4\cdot\left(-2\right)}=\frac{4z\cdot3z}{3\cdot3}=\frac{6x}{10}=\frac{10y}{-8}=\frac{12z}{9}=\frac{6x-10y+12z}{10-\left(-8\right)+9}=\frac{30}{27}=\frac{10}{9}\)

\(\frac{6x}{10}=\frac{10}{9}\Rightarrow x=\frac{10\cdot10}{9}:6=1\frac{23}{27}\)

\(\frac{10y}{-8}=\frac{10}{9}\Rightarrow y=\frac{10\cdot\left(-8\right)}{9}:10=-\frac{8}{9}\)

\(\frac{12z}{9}=\frac{10}{9}\Rightarrow z=\frac{9\cdot10}{9}:12=\frac{5}{6}\)

\(15x=-10y=6z\)  =>  \(\frac{x}{2}=\frac{y}{-3}=\frac{z}{5}=k\)

=>  \(x=2k;\) \(y=-3k;\)\(z=5k\)

Ta có: \(x.y.z=-30000\)

<=> \(2k.\left(-3k\right).5k=-30000\)

<=> \(-30k^3=30000\)

<=> \(k^3=1000\)

<=> \(k=10\)

suy ra: \(x=20;\)\(y=-30\)\(z=50\)

\(15x=-10y=6z\Leftrightarrow\frac{15x}{30}=\frac{-10y}{30}=\frac{6z}{30}\)

\(\Leftrightarrow\frac{x}{2}=\frac{-y}{3}=\frac{1}{5}\)

Đặt \(\frac{x}{2}=\frac{-y}{3}=\frac{1}{5}=k\)

\(\Rightarrow x=2k;y=-3k;z=5k\)

\(\Rightarrow xyz=-30000\Leftrightarrow2k\cdot\left(-3\right)k\cdot5k=-30000\)

\(\Leftrightarrow-30k^3=-30000\)

\(\Leftrightarrow k^3=1000\)\(\Leftrightarrow k=10\)

\(\Rightarrow\begin{cases}x=2k=2\cdot10=20\\y=\left(-3\right)k=-3\cdot10=-30\\z=5k=5\cdot10=50\end{cases}\)

Ta có: 15x = -10y = 6z

\(=\frac{x}{\frac{1}{15}}=\frac{y}{\frac{-1}{10}}=\frac{z}{\frac{1}{6}}\)

\(\Rightarrow\left(\frac{x}{\frac{1}{15}}\right)^3=\left(\frac{y}{\frac{-1}{10}}\right)^3=\left(\frac{z}{\frac{1}{6}}\right)^3=\frac{x}{\frac{1}{15}}.\frac{y}{\frac{-1}{10}}.\frac{z}{\frac{1}{6}}\)

\(=\frac{x^3}{\frac{1}{15^3}}=\frac{y^3}{\frac{-1}{10^3}}=\frac{z^3}{\frac{1}{6^3}}=\frac{-30000}{\frac{-1}{900}}=300\)

\(\Rightarrow\begin{cases}x^3=300^3.\frac{1}{15^3}=20^3\\y^3=300^3.\frac{-1}{10^3}=-30^3\\z^3=300^3.\frac{1}{6^3}=50^3\end{cases}\)\(\Rightarrow\begin{cases}x=20\\y=-30\\z=50\end{cases}\)

Vậy x = 20; y = -30; z = 50

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