\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\cdot\left(x+1\right)}=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\cdot\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2013}{2015}:2\)

\(\Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\)

\(\Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\Rightarrow x+1=2015\Rightarrow x=2014\)

bạn trên trả lời đúng rùi đó

Trần Thị Loan chuẩn rồi 

= 2/(2.3) + 2/3.4 + 2/4.5 +...+ 2/x(x+1)

= 2 [1/2-1/3+1/3-1/4+...+1/x-1/(x+1)]

=2[1/2-1/(x+1)]= (x-1)/(x+1)

= 2001/2003

==> x=2002

x=2002

Mình Giúp Họ Giải Toán Đầu tiên Mà Họ Lại Làm Ngơ sai bét

nhân cả 2 vế của đẳng thức với 1/2 ta được

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}=\frac{2014}{2015}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-......+\frac{1}{x}-\frac{1}{x+1}=\frac{2014}{2015}\)

\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2014}{2015}\)

\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2014}{2015}\)

        \(\frac{1}{x+1}=-\frac{2013}{4030}\)

hay \(1:\left(x+1\right)=-\frac{2013}{4030}\)

       \(x+1=-\frac{4030}{2013}\)

\(=>x=-\frac{6043}{2013}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2013}{2015}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{4030}\)

tự làm tiếp nhé mk ăn cơm đã