\(15x=10y=6z\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{20}{10}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\\z=2.5=10\end{matrix}\right.\)

Vì 35x=14y=10z
=> \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có : 
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{\left(x+z-y\right)}{2+5-7}=\dfrac{20}{0}=0\)

Có : x/2 = 0 => x = 2*0 = 0
        y/5 = 0 => y = 5*0 = 0
        z/7 = 0 => z=7*0=0
Vậy, ..

a)

\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y+1}=\frac{x+y+z}{2\left(x+y+z\right)+3}=x+y+z\)

=> 2(x+y+z) +3 =1=> x+y+z=-1

Luôn đùng Vói mọi x;y;z khác o  sao cho x+y+z = -1

b)\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}=\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)

x= 3/2 .12=18

y= 4/3 .12=16

z=5/4 .12=15

6/11.x = 9/2.y = 15/5.z 6/11.x = 9/2.y = 3.z =) 6/11.x = 9/2.(-3)=(-y).z =) 6/11.x = (-27/2) = (-y).z • x = (-27/2) : 6/11 x = -99/4 ( mk làm tắt) Vì x = (-99/4) =) 6/11.(-99/4)= 9/2.y= 3.z =) -27/2=9/2.y=3.z •9/2.y=-27/2 y=-27/2:9/2 y=-3 •3.z=-27/2 z=-27/2:3 z= -9/2 Vậy x=-99/4;y=-3;z=-9/2 (Mk làm bừa câu này còn câu còn lại ko biết)

Theo đề: \(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow-\left(x+y\right)=z\)

\(\Leftrightarrow-\left(x+y\right)^5=z^5\)

\(x^2+y^2+z^2=1\)

\(\Rightarrow x^2+y^2=1-z^2\)

\(\Rightarrow\left(x+y\right)^2-2xy=1-z^2\)

\(\Rightarrow\left(x+y\right)^2=1-z^2+2xy\)

\(\Rightarrow\left(-z\right)^2=1-z^2+2xy\)

\(\Leftrightarrow xy=\frac{2z^2-1}{2}\)

Nên ta có:

\(VT=x^5+y^5+z^5=x^5+y^5-\left(x+y\right)^5\)

                                   \(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)

                                   \(=x^5+y^5-x^5-5x^4y-10x^3y^2-10x^2y^3-5xy^4-y^5\)

                                    \(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)

                                    \(=-5xy\left(x^3+y^3\right)-10x^2y^2\left(x+y\right)\)

                                    \(=-5xy\left(x+y\right)\left(x^2-xy+y^2\right)-10x^2y^2\left(x+y\right)\)

                                     \(=-5xy\left(x+y\right)\left(x^2-xy+y^2+2xy\right)\)

                                     \(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)

                                       \(=-5.\frac{2z^2-1}{2}.\left(-z\right).\left(1-z^2+\frac{2z^2-1}{2}\right)\)

                                       \(=\frac{5z\left(2z^2-z\right)}{4}=\frac{5}{4}z\left(2x^2-1\right)=\frac{5}{4}\left(2z^3-z\right)=VP\)

=> đpcm

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{4}=\frac{z}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{y}{12}\\\frac{y}{12}=\frac{z}{15}\end{cases}}\)\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}\)

\(\Rightarrow\frac{x^2}{64}=\frac{y^2}{144}=\frac{z^2}{225}=\frac{x^2-y^2}{64-144}=\frac{-16}{-80}=\frac{1}{5}\)

\(\Rightarrow\hept{\begin{cases}x^2=\frac{1}{5}.64=12,8\\y^2=\frac{1}{5}.144=28,8\\z^2=\frac{1}{5}.225=45\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\pm\sqrt{12,8}\\y=\pm\sqrt{28,8}\\z=\pm\sqrt{45}\end{cases}}\)

Với \(x=\sqrt{12,8}\Rightarrow\hept{\begin{cases}y=\sqrt{28,8}\\z=\sqrt{45}\end{cases}}\)

Với \(x=-\sqrt{12,8}\Rightarrow\hept{\begin{cases}y=-\sqrt{28,8}\\z=-\sqrt{45}\end{cases}}\)