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8 tháng 7 2017

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13 tháng 8 2016
  • Ta có : x + (x+1) + (x+2) + ... + (x+2010) = 2029099

=> 2011x + 1 + 2 +... + 2010 = 2029099

=> \(2011x+\frac{2010.2011}{2}=2029099\)

\(\Rightarrow2011x=2029099-2021055=8044\)

=> x = 4

  • 2 + 4 + 6 + 8 +... + 2x = 210

=> 2.1+2.2+2.3+2.4+...+2.x=210

=> 2(1+2+3+...+x) = 210

=> 1 + 2 +3 +...+ x = 105

=> \(\frac{x\left(x+1\right)}{2}=105\Leftrightarrow x\left(x+1\right)=210=14.15\)

=> x = 14

a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)

\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)

=>2x+10=0

hay x=-5

b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)

=>x-2017=0

hay x=2017

a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)

\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)

=>2x+10=0

hay x=-5

b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)

=>x-2017=0

hay x=2017