Nhẩm nghiệm ta thấy: a+b+c=3 \(\Rightarrow\)a=b=c=1    (1)

Áp dụng bất đẳng thức AM-GM, ta có:

\(x^5+y^5+z^5+\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\ge6\sqrt[6]{\frac{x^5y^5z^5}{xyz}}=6\sqrt[6]{x^4y^4z^4}\)

Hay: \(6\sqrt[6]{x^4y^4z^4}\ge6\)

\(\Leftrightarrow\sqrt[6]{x^4y^4z^4}=1\Leftrightarrow x^4y^4z^4=1\Leftrightarrow xyz=1\)   (2)

Từ (1) và (2) suy ra: x=y=z=1

\(\frac{2.\left(x+y\right)}{30}=\frac{5.\left(y+z\right)}{30}=\frac{3\left(x+z\right)}{30}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)=\(\frac{x+z-y-z}{10-6}=\frac{x-y}{4}\)

\(\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)=\(\frac{x+y-x-z}{15-10}=\frac{y-z}{5}\)

---> dp cm

Theo đề ta có:

\(2\left(x+y\right)=5\left(y+z\right)=3\left(x+z\right)\)

\(\Rightarrow\)\(\frac{2\left(x+y\right)}{30}=\frac{5\left(y+z\right)}{30}=\frac{3\left(x+z\right)}{30}\)

\(\Rightarrow\frac{x+y}{15}=\frac{y+z}{6}=\frac{x+z}{10}\)

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

* \(\frac{y+z}{6}=\frac{x+z}{10}=\frac{\left(x+z\right)-\left(y+z\right)}{10-6}=\frac{x+z-y-z}{4}=\frac{x-y}{4}\)  \(\left(1\right)\)

* \(\frac{x+y}{15}=\frac{x+z}{10}=\frac{x+y-\left(x+z\right)}{15-10}=\frac{x+y-x-z}{5}=\frac{y-z}{5}\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)ta có:

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

a./ \(\frac{x}{5}=\frac{y}{7}=\frac{z}{4}=\frac{x-y+z}{5-7+4}=\frac{-10}{2}=-5\)

\(\Rightarrow x=-25;y=-35;z=-20\)

b./ \(\frac{x}{5}=\frac{y}{-4}=\frac{z}{-7}=\frac{x+y-z}{5-4-\left(-7\right)}=\frac{-40}{6}=-5\)

\(\Rightarrow x=-25;y=20;z=35\)

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

a./ \(\frac{x}{5}=\frac{y}{4}=\frac{z}{7}=\frac{2y}{8}=\frac{x+2y+z}{5+8+7}=\frac{10}{20}=\frac{1}{2}\)

\(\Rightarrow x=\frac{5}{2};y=2;z=\frac{7}{2}\)

b./ \(\frac{x}{4}=\frac{y}{5}=\frac{z}{2}=\frac{x+y}{9}=\frac{18}{9}=2\)

\(\Rightarrow x=2\cdot4=8;y=2\cdot5=10;z=2\cdot2=4\)