Tính B=1+\(\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
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Những câu hỏi liên quan
7 tháng 12 2021
\(B=1+\dfrac{1}{2}\cdot\dfrac{\left(1+2\right)\cdot2}{2}+\dfrac{1}{3}\cdot\dfrac{\left(1+3\right)\cdot3}{2}+...+\dfrac{1}{20}\cdot\dfrac{\left(20+1\right)\cdot20}{2}\\ B=1+\dfrac{3}{2}+2+\dfrac{5}{2}+...+10+\dfrac{21}{2}\\ B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{20}{2}+\dfrac{21}{2}\\ B=\dfrac{2+3+...+20+21}{2}=\dfrac{\dfrac{\left(21+2\right)\cdot20}{2}}{2}=\dfrac{23\cdot10}{2}=115\)
10 tháng 3 2017
\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)
\(\Rightarrow B=1+\dfrac{1}{2}.2.3\div2+\dfrac{1}{3}.3.4\div2+...+\dfrac{1}{20}.20.21\div2\)
\(\Rightarrow B=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}\)
\(\Rightarrow B=\dfrac{2+3+4+...+21}{2}\)
\(\Rightarrow B=\dfrac{230}{2}\)
\(\Rightarrow B=115\)
Vậy \(B=115\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 11 2023
A = - 522 - { - 222 - [ - 122 - (100 - 522) + 2022] }
A = - 522 - { -222 - [- 122 - 100 + 522 ] + 2022}
A = - 522 - { -222 - { - 222 + 522 } + 2022}
A = - 522 - {- 222 + 222 - 522 + 2022}
A = -522 + 522 - 2022
A = - 2022
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
18 tháng 11 2023
B = 1 + \(\dfrac{1}{2}\)(1 + 2) + \(\dfrac{1}{3}\).(1 + 2 + 3) + ... + \(\dfrac{1}{20}\).(1 + 2+ 3 + ... + 20)
B = 1+\(\dfrac{1}{2}\)\(\times\)(1+2)\(\times\)[(2-1):1+1]:2+ ... + \(\dfrac{1}{20}\)\(\times\) (20 + 1)\(\times\)[(20-1):1+1]:2
B = 1 + \(\dfrac{1}{2}\) \(\times\) 3 \(\times\) 2:2 + \(\dfrac{1}{3}\) \(\times\)4 \(\times\) 3 : 2+....+ \(\dfrac{1}{20}\) \(\times\)21 \(\times\) 20 : 2
B = 1 + \(\dfrac{3}{2}\) + \(\dfrac{4}{2}\) + ....+ \(\dfrac{21}{2}\)
B = \(\dfrac{2+3+4+...+21}{2}\)
B = \(\dfrac{\left(21+2\right)\left[\left(21-2\right):1+1\right]:2}{2}\)
B = \(\dfrac{23\times20:2}{2}\)
B = \(\dfrac{23\times10}{2}\)
B = 23
20 tháng 12 2021
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
30 tháng 1 2023
\(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)
\(=1+\dfrac{3\cdot2\div2}{2}+\dfrac{4\cdot3\div2}{3}+...+\dfrac{21\cdot20\div2}{20}\)
\(=1+\dfrac{3}{2}+2+...+\dfrac{21}{2}\) (A)
Trong (A) có \(\dfrac{\dfrac{21}{2}-1}{\dfrac{3}{2}-1}+1=20\) (số hạng)
Suy ra \(\left(A\right)=\left(\dfrac{21}{2}+1\right)\cdot20\div2=115\)
Vậy \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)=115\)
TH
4 tháng 12 2018
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
10 tháng 12 2023
Sửa đề:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)
PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)
=>\(x-1=\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)