\(\frac{x}{y}=\frac{x}{5};\frac{x}{y}=\frac{y}{9};\frac{z}{2}=\frac{3}{8}\)

=> \(\frac{x}{5}=\frac{y}{9};\frac{z}{2}=\frac{3}{8}\)

*) \(\frac{x}{5}=\frac{y}{9}\Rightarrow\)\(\frac{x}{y}=\frac{5}{9}\)

*) \(\frac{z}{2}=\frac{3}{8}\Rightarrow z=\frac{2\cdot3}{8}=\frac{3}{4}\)

++) \(\frac{x}{y}=\frac{5}{9}\) và \(\frac{x}{y}=\frac{x}{5}\Rightarrow\)\(\frac{5}{9}=\frac{x}{5}\Rightarrow\)\(x=\frac{5\cdot5}{9}=\frac{25}{9}\)

++) \(\frac{x}{y}=\frac{5}{9}\) và \(\frac{x}{y}=\frac{y}{9}\Rightarrow\) \(\frac{5}{9}=\frac{y}{9}\Rightarrow y=\frac{5\cdot9}{9}=5\)

Vậy x = 25/9 , y = 5, z = 3/4

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{100-1}{100!}=..........=1-\frac{1}{100!}

\(\frac{1}{2!}\)+ \(\frac{2}{3!}\)+ \(\frac{3}{4!}\)+ ... + \(\frac{99}{100!}\)

= \(\frac{2-1}{2!}\)+ \(\frac{3-1}{3!}\)+ ... + \(\frac{100-1}{100!}\)

= \(1\)\(-\)\(\frac{1}{100!}\)\(< \)\(1\)\(\left(đpcm\right)\)

\(\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}=\left(2004-1-1-...-1\right)+\left(\frac{2003}{2}+1\right)+...+\left(\frac{1}{2004}+1\right)\)

\(=1+\frac{2005}{2}+...+\frac{2005}{2014}=2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2004}\right)\)

vậy \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{\frac{2004}{1}+\frac{2003}{2}+...+\frac{1}{2004}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}}{2005\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2005}\right)}=\frac{1}{2005}\)

Ta có:

\(\frac{a}{2}=\frac{b}{3};\frac{b}{5}=\frac{c}{4}\Rightarrow\)\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}\)

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

\(\frac{a}{10}=\frac{b}{15}=\frac{c}{12}=\frac{a-b+c}{10-15+12}=\frac{-49}{7}=-7\)

\(\frac{a}{10}=-7\Rightarrow a=-7\cdot10=-70\)

\(\frac{b}{15}=-7\Rightarrow b=-7\cdot15=-105\)

\(\frac{c}{12}=-7\Rightarrow c=-7\cdot12=-84\)

Vậy a = -70, b = -105, c = -84

\(\frac{a}{3b}=\frac{b}{3c}=\frac{c}{3d}=\frac{d}{3a}=\frac{a+b+c+d}{3a+3b+3c+3d}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)

\(\Rightarrow\frac{a}{3b}=\frac{a}{3a}\Rightarrow3b=3a\Rightarrow a=b\)

\(\frac{b}{3c}=\frac{b}{3b}\Rightarrow3b=3c\Rightarrow b=c\)

\(\frac{c}{3d}=\frac{c}{3c}\Rightarrow3c=3d\Rightarrow c=d\)

=>a=b=c=d

=>đpcm

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

Suy ra: \(\frac{1}{x+y+z}=2\Rightarrow x+y+z=\frac{1}{2}\)(*)

Ta có: \(\frac{y+z+1}{x}=2\Leftrightarrow y+z+1=2x\Leftrightarrow x+y+z+1=3x\Leftrightarrow\frac{1}{2}+1=3x\Leftrightarrow x=\frac{1}{2}\)

\(\frac{x+z+2}{y}=2\Leftrightarrow x+z+2=2y\Leftrightarrow x+y+z+2=3y\Leftrightarrow\frac{1}{2}+2=3y\Leftrightarrow y=\frac{5}{6}\)

Từ (*) suy ra: \(z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}\Leftrightarrow z=-\frac{5}{6}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}\)

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

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

Ta có : \(\frac{y+z+1}{x}=2\Leftrightarrow y+z+1=2x\Rightarrow x+y+z+1=3x\Rightarrow\frac{1}{2}+1=3x\Leftrightarrow x=\frac{1}{2}\)

\(\frac{x+z+2}{y}=2\Leftrightarrow x+z+2=2y\Leftrightarrow x+y+z+2=3y\Leftrightarrow\frac{1}{2}+2=3y\Leftrightarrow y=\frac{5}{6}\)

Từ \(\left(\cdot\right)\Rightarrow z=\frac{1}{2}-x-y=\frac{1}{2}-\frac{1}{2}-\frac{5}{6}\Leftrightarrow z=-\frac{5}{6}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}\)