Ai làm đc giải cụ thể ra nha!thanks

<=> 1/3 + 1/6 + 1/10 +...+ 1/x(x+1):2 = 1/1991/1993 - 1 = 1991/1993

<=> 1/2(2+1):2 + 1/3(3+1):2 + ...+ 1/x(x+1):2 = 1991/1993

<=> 1/2.3:2 + 1/3.4:2 +...+ 1/x(x+1):2 = 1991/1993

<=>(1/2 - 1/3):1/2 + (1/3 - 1/4 ):1/2+...+(1/x-1/x+1):1/2=1991/1993

<=>(1/2-1/3).2 + (1/3-1/4).2+...+(1/x-1/x+1).2 = 1991/1993

<=>2.(1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/x+1)=1991/1993

<=>2.(1/2-1/x+1)=1991/1993

<=>1/2-1/x+1=1991/1993:2=1991/3986

<=> 1/x+1=1/2-1991/3986=2/3986=1/1993

=>x=1993-1=1992

<=> 1/3 + 1/6 + 1/10 +...+ 1/x(x+1):2 = 1/1991/1993 - 1 = 1991/1993

<=> 1/2(2+1):2 + 1/3(3+1):2 + ...+ 1/x(x+1):2 = 1991/1993

<=> 1/2.3:2 + 1/3.4:2 +...+ 1/x(x+1):2 = 1991/1993

<=>(1/2 - 1/3):1/2 + (1/3 - 1/4 ):1/2+...+(1/x-1/x+1):1/2=1991/1993

<=>(1/2-1/3).2 + (1/3-1/4).2+...+(1/x-1/x+1).2 = 1991/1993

<=>2.(1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/x+1)=1991/1993

<=>2.(1/2-1/x+1)=1991/1993

<=>1/2-1/x+1=1991/1993:2=1991/3986

<=> 1/x+1=1/2-1991/3986=2/3986=1/1993

=>x=1993-1=1992

1+13+16+110+...+1x(x+2):2=19911993

⟹12+16+112+120+...+1x(x+2)=19911993.2

⟹11.2+12.3+13.4+14.5+...+1x.(x+2)=19911993.2

⟹1−12+12−13+14−15+...+1x−1x+2=19911993.2

⟹1−1x+2=19913986

⟹1x+2=19913986−1

⟹1x+2=−19953986

⟹x+2=−39861995

⟹x=−79761995
 

<=> 1/3 + 1/6 + 1/10 +...+ 1/x(x+1):2 = 1/1991/1993 - 1 = 1991/1993

<=> 1/2(2+1):2 + 1/3(3+1):2 + ...+ 1/x(x+1):2 = 1991/1993

<=> 1/2.3:2 + 1/3.4:2 +...+ 1/x(x+1):2 = 1991/1993

<=>(1/2 - 1/3):1/2 + (1/3 - 1/4 ):1/2+...+(1/x-1/x+1):1/2=1991/1993

<=>(1/2-1/3).2 + (1/3-1/4).2+...+(1/x-1/x+1).2 = 1991/1993

<=>2.(1/2-1/3+1/3-1/4+1/4-1/5+....+1/x-1/x+1)=1991/1993

<=>2.(1/2-1/x+1)=1991/1993

<=>1/2-1/x+1=1991/1993:2=1991/3986

<=> 1/x+1=1/2-1991/3986=2/3986=1/1993

=>x=1993-1=1992

1+13+16+110+...+1x(x+2):2=19911993

⟹12+16+112+120+...+1x(x+2)=19911993.2

⟹11.2+12.3+13.4+14.5+...+1x.(x+2)=19911993.2

⟹1−12+12−13+14−15+...+1x−1x+2=19911993.2

⟹1−1x+2=19913986

⟹1x+2=19913986−1

⟹1x+2=−19953986

⟹x+2=−39861995

⟹x=−79761995

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

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{3984}{1993}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{3984}{1993}\div2\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1992}{1993}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1992}{1993}\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)

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

Vây x = 1992