\(\Leftrightarrow\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{x\left(x+1\right):2}=\dfrac{1991}{1993}\)

\(\Leftrightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{1991}{1993}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1991}{1993}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1991}{3986}\)

=>1/x+1=1/1993

=>x+1=1993

hay x=1992

\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{1991}{1993}\)

\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)

\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)

\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1991}{1993}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)\(\Rightarrow\frac{1}{x+1}=\frac{1}{1993}\)

\(\Rightarrow x+1=1993\Rightarrow x=1992\)

=> 1/2+1/6+1/12+1/20+....+1/x.(x+1) = 1992/1993

=> 1/2+1/2.3+1/3.4+1/4.5+.....+1/x.(x+1) = 1992/1993

=> 1/2+1/2-1/3+1/3-1/4+1/4-1/5+.....+1/x-1/x+1 = 1992/1993

=> 1 - 1/x+1 = 1992/1993

=> x/x+1 = 1992/1993

=> x = 1992

Vậy x = 1992

Tk mk nha

\(\Rightarrow\frac{2}{2}+\frac{2}{2.3}+\frac{2}{2.6}+...+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)

\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{3984}{1993}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3984}{1993}:2\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Rightarrow\frac{x}{x+1}=\frac{1992}{1993}\)

=>x=1992

mina toán 6 sau khó vậy bó tay .com.vn

chuyển 1 đi còn cái kia=1991/1993

nhân mỗi p/số với 1/2 rồi p/tích mẫu=2.3,3.4........x.(x+1)

lập hiệu ra rồi tính OK???

=2/2+2/6+2/12+...+...+2/x(x+1)=3984/1993

=2(1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=3984/1993

=2(1-1/2+1/2-1/3+14-1/5+...+1/x-1/x+1=3984/1993

=2(1-1/x+1)=3984/1993

=2-2/x+1=3984/1993

2/x+1=2/1993

x+1=1003

x=1992

=2/2+2/6+2/12+...+...+2/x(x+1)=3984/1993

=2(1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=3984/1993

=2(1-1/2+1/2-1/3+14-1/5+...+1/x-1/x+1=3984/1993

=2(1-1/x+1)=3984/1993

=2-2/x+1=3984/1993

2/x+1=2/1993

x+1=1003

x=1992

\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)

=> \(\frac{1}{1.\left(1+1\right):2}+\frac{1}{2.\left(2+1\right):2}+\frac{1}{3.\left(3+1\right):2}+\frac{1}{4.\left(4+1\right):2}+...+\frac{1}{x.\left(x+1\right):2}=1\frac{1991}{1993}\)

=> \(\frac{1}{\frac{1.\left(1+1\right)}{2}}+\frac{1}{\frac{2.\left(2+1\right)}{2}}+...+\frac{1}{\frac{x.\left(x+1\right)}{2}}=1\frac{1991}{1993}\)

=> \(\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{x.\left(x+1\right)}=1\frac{1991}{1993}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=1\frac{1991}{1993}\)

=> \(1-\frac{1}{x+1}=1\frac{1991}{1993}\)

=> \(\frac{1}{x+1}=\frac{-1991}{1993}\)

=> -1991.(x + 1) = 1993

=> -1991x - 1991 = 1993

=> -1991x = 3984

=> x = \(-\frac{3984}{1991}\)