\(x=15,51\)

Gọi ba số cần tìm là a-1, a, a+1 
ta có : 
(a-1)a+50=(a+1)a 
a^2-a+50=a^2+a 
a^2-a^2+50=a+a 
suy ra 2a=50 
a=50/2=25 
a-1=25-1=24 
a+1=25+1=26 
Vậy ba số cần tìm là : 24,25,26

x<-1/4 <=> E=-3x+7-4x-1=-7x+6

\(=\frac{-37}{38}\cdot\frac{-36}{37}\cdot...\cdot\frac{-1}{2}=\frac{\left(-37\right)\left(-36\right)\cdot...\cdot\left(-1\right)}{38\cdot37\cdot...\cdot2}=\frac{1}{38}\)

\(\frac{1}{5.6}\)- \(\frac{1}{6.7}\)- \(\frac{1}{7.8}\) - ... - \(\frac{1}{2004.2005}\) 

= \(\frac{1}{5}\)- \(\frac{1}{6}\)+ \(\frac{1}{6}\)- \(\frac{1}{7}\)+ \(\frac{1}{7}\)- \(\frac{1}{8}\)+ ... + \(\frac{1}{2004}\)- \(\frac{1}{2005}\)

=\(\frac{1}{5}\)- \(\frac{1}{2005}\) 

=  \(\frac{80}{401}\)

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{a\left(bz-cy\right)}{a^2}=\frac{b\left(cx-az\right)}{b^2}=\frac{c\left(ay-bx\right)}{c^2}\)

=\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

=\(\frac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0\)

suy ra \(\frac{bz-cy}{a}=0\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{z}{c}=\frac{y}{b}\left(1\right)\)

\(\frac{cx-az}{b}=0\Rightarrow cx-az=0\Rightarrow cx=az\Rightarrow\frac{x}{a}=\frac{z}{c}\left(2\right)\)

từ (1) và (2) suy ra \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

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