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\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+...+\frac{2}{x\left(x+1\right)}=\frac{2010}{2012}\)
\(2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)
\(2\left(\frac{1}{4}-\frac{1}{x+1}\right)=\frac{2010}{2012}\)
\(\frac{1}{4}-\frac{1}{x+1}=\frac{2010}{2012}\div2\)
\(\frac{1}{4}-\frac{1}{x+1}=\frac{1005}{2012}\)
\(\frac{1}{x+1}=\frac{1}{4}-\frac{1005}{2012}\)
\(\frac{1}{x+1}=\frac{-502}{2012}=-\frac{251}{1006}\)
\(\Rightarrow x+1=1\div-\frac{251}{1006}=-\frac{1006}{251}\)
\(x=\frac{-1006}{251}-1=-\frac{1257}{251}\)
a)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+2010\right)=2029099\\ 2011.x+\left(1+2+3+...+2010\right)=2029099\\ 2011.x+2021055=2029099\\ 2011.x=2029099-2021055\\ 2011.x=8044\\ x=8044:2011\\ x=4\)
b)
\(2+4+6+...+2x=210\\ 2.\left(1+2+3+...+x\right)=210\\ 1+2+3+...+x=210:2\\ 1+2+3+...+x=105\\ \dfrac{x.\left(x+1\right)}{2}=105\\ x.\left(x+1\right)=105.2\\ x\left(x+1\right)=210\\ x.\left(x+1\right)=14.15\\\Rightarrow x=14\)
a) x+(x+1)+(x+2)+...+(x+2010)=2029099
x+x+1+x+2+...+x+2010=2029099
2011x+[(2010+1).2010:2]=2029099
2011x+2021055=2029099
2011x=2029099-2021055
2011x=8044
x=8044:2011
x=4
Vậy x=4.
b) 2+4+6+8+...+2x=210
(2x+2)*14:2=210
(2x+2)*7=210
2x+2=210:7
2x+2=30
2x=30-2
2x=28
x=28:2
x=14
Vậy x=14.
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