\(\frac{5-x}{-25-y}=\frac{x}{y}\)

\(\Leftrightarrow y\left(5-x\right)=x\left(-25-y\right)\)

\(\Leftrightarrow5y-xy=-25y-xy\)

\(\Leftrightarrow5x=-25y\)

\(\Rightarrow\frac{y}{x}=\frac{5}{-25}=-\frac{1}{5}\)

Hùng sai òi : 

Ta có ; \(\frac{5-x}{-25-y}=\frac{x}{y}\)

\(\Rightarrow\left(5-x\right)y=\left(-25-y\right)x\)

\(\Rightarrow5y-xy=-25x-xy\)

\(\Rightarrow5y=-25x\)

Vậy \(\frac{x}{y}=\frac{5}{-25}=\frac{-1}{5}\)

\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)

=> (5x - 2y).4 = 7.(x + 3y)

=> 20x - 8y = 7x + 21y

=>> 20x - 7x = 21y + 8y

=> 13x = 29y

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

\(\frac{5x-2y}{x+3y}=\frac{7}{4}\)

\(\Rightarrow4\left(5x-2y\right)=7\left(x+3y\right)\)

\(\Rightarrow20x-8y=7x+21y\)

\(\Rightarrow20x-7x=8y+21y\)

\(\Rightarrow13x=29y\)

\(\Rightarrow\frac{x}{y}=\frac{29}{13}\)

Vậy \(\frac{x}{y}=\frac{29}{13}\)

\(2x^3-1=15\Rightarrow x^3=\frac{15+1}{2}=8\Rightarrow x=2\)

Thay x = 2, ta được

\(\frac{2+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=2\)

\(y-25=2.16=32\Rightarrow y=57\)

\(z+9=2.25=50\Rightarrow z=41\)

\(x+y+z=\)\(2+57+41\)\(=100\)

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

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)

Do đó : 

\(\frac{y+z-x}{x}=1\)\(\Rightarrow\)\(2x=y+z\)

\(\frac{z+x-y}{y}=1\)\(\Rightarrow\)\(2y=x+z\)

\(\frac{x+y-z}{z}=1\)\(\Rightarrow\)\(2z=x+y\)

Suy ra : 

\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{x}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)

Vậy \(P=8\)

Đề hơi sai