Question

[Thân Thị Thảo Quỳnh]

Tìm x

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

Answer 1

[I don't need you]

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+1+(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013})=2\)

\(\frac{1}{2013}x+1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+1+\left(1-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+1+1-\frac{1}{2013}=2\)

\(\frac{1}{2013}x-\frac{1}{2013}+2=2\)

\(\frac{1}{2013}.\left(x-1\right)=2-2\)

\(\frac{1}{2013}.\left(x-1\right)=0\)

=> x - 1 = 0

x = 1

Answer 2

[Nhũn]

\(\frac{1}{2013}x+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2012.2013}=2\)

\(\frac{1}{2013}x+\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)=2\)

\(\frac{1}{2013}x+\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+\left(1-\frac{1}{2013}\right)=2\)

\(\frac{1}{2013}x+\frac{2012}{2013}=2\)

\(\frac{1}{2013}x=2-\frac{2012}{2013}\)

\(\frac{1}{2013}x=\frac{2014}{2013}\)

\(x=\frac{2014}{2013}:\frac{1}{2013}\)

=> x=2014