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18 tháng 9 2021
Nguồn: Tính tổng: 1x2 + 2x3 + 3x4 +...+ 2019x2020 + 2020x2021 - Hoc24
Đặt
NC
23 tháng 8 2020
Bài làm:
Ta có: \(\frac{x}{1.2}+\frac{x}{2.3}+\frac{x}{3.4}+...+\frac{x}{2017.2018}=-1\)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2018}\right)=-1\)
\(\Leftrightarrow x.\frac{2017}{2018}=-1\)
\(\Rightarrow x=-\frac{2018}{2017}\)
CB
2
HT
25 tháng 7 2015
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(1-\frac{1}{x+1}=\frac{99}{100}\)
=> \(\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)
=> x+1 = 100
=> x = 100 - 1
=> x = 99
KS
7 tháng 4 2022
1/1.2 +1/2.3 +1/3.4 +....+1/99.100
=1-1/2+1/2-1/3+1/3-14+.....+1/99-1/100
=1-1/100
=99/100
LN
1
25 tháng 10 2021
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}=\dfrac{2009}{2010}\)
PA
0
4 tháng 7 2019
\(\frac{1313}{1212}:x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=1\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
4 tháng 7 2019
Lời giải
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}\)
\(\frac{1313}{1212}:x=\frac{4}{5}+\frac{1}{5}\)
\(x=\frac{1313}{1212}:1\)
\(x=\frac{13}{12}\)
13 tháng 7 2023
\(C=1.2+2.3+3.4+...+x.\left(x-1\right)\)
\(\Rightarrow3C=1.2.3+2.3.3+3.4.3+...+x.\left(x-1\right).3\)
\(\Rightarrow3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+x.\left(x-1\right).\left[\left(x+1\right)-\left(x-2\right)\right]\)
\(\Rightarrow3C=\left(1.2.3-0.12\right)+\left(2.3.4-1.2.3\right)+\left(3.4.5-2.3.4\right)+...+\left[x.\left(x-1\right)\left(x+1\right)-x.\left(x-1\right)\left(x-2\right)\right]\)
\(\Rightarrow3C=-0.1.2+x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow3C=x.\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow C=\dfrac{x.\left(x-1\right)\left(x+1\right)}{3}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 7 2023
3C=1x2x3+2x3x3+3x4x3+...+Xx(X+1)=
=1x2x3+2x3x(4-1)+3x4x(5-2)+...+Xx(X+1)[(X+2)-(X-1)]=
=1x2x3-1x2x3+2x3x4-2x3x4+3x4x5-...-(X-1)xXx(X+1)+Xx(X+1)x(X+2)=
=Xx(X+1)(X+2)
\(\Leftrightarrow x\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}\right)=-1\)
\(\Leftrightarrow x\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{2021-2020}{2020.2021}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\right)=-1\)
\(\Leftrightarrow x\left(1-\frac{1}{2021}\right)=-1\Leftrightarrow x.\frac{2020}{2021}=-1\Rightarrow x=-\frac{2021}{2020}\)