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<=> 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
<=> 1/3 + 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
=> 1/2+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
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