\(=\dfrac{1+2+3+...+99+100}{5\left(1+2+3+4+...+99+100\right)}=\dfrac{1}{5}\)
Đúng 2
Bình luận (0)
Các câu hỏi tương tự
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
Tính A=92-1/9-2/10-3/11-4/12-5/3-...-91/99-92/100 / 1/45+1/50+1/55+...+1/495+1/500
9 tháng 5 2021 lúc 20:08
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
94-\(\dfrac{1}{7}-\dfrac{2}{8}-\dfrac{3}{9}-...-\dfrac{94}{100}\)
\(\dfrac{1}{35}+\dfrac{1}{40}+\dfrac{1}{45}+...+\dfrac{1}{500}\)
giúp mik nha mik cần gấp
Tính nhanh các tổng sau
a)11+12+13+14+15+16+17+18+19 b)1+2+3+4+5+................+99+100
c)2+4+6+8+10+.........100 d)5+10+15+20+25+......+95+100
GIÚP MÌNH VỚI !
A=\(\dfrac{1}{2}\) nhân \(\dfrac{3}{4}\)nhân \(\dfrac{5}{6}\)nhân ..... nhân \(\dfrac{99}{100}\)và B=\(\dfrac{1}{10}\)
a,-5+(-10)+(-15)+(-20)+.......+(-100)
b,-4+(-8)+(-12)+(-16)+.........+(-100)
c,1+(-3)5+(-7)+9+(-11)+.......+97+(-99)
25 tháng 3 2022 lúc 19:55
Cho A = \(\dfrac{1}{2}x\dfrac{3}{4}x\dfrac{5}{6}x...x\dfrac{99}{100};B=\dfrac{1}{10}\) So sánh: A và B.
1.tính
a)1-2+3-4+5-6+7-8+8-9+9-10
b)1-2+3-4+...+99-100
c)1-3+5-7+9-11+13-15
d)1-3+5-7+...+99-101
e)-1-2-3-4-...-99-100
20^5-5^10:100^5
B=3^100-3^99+3^98-.........+3^2-3+1
D=2^100-2^99+2^98-.........+2^2-2+1
11 tháng 4 2021 lúc 15:06
text{Bài 4. Chứng tỏ rằng:}a) dfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{30^2} 1b) dfrac{1}{10}+dfrac{1}{11}+dfrac{1}{12}+...+dfrac{1}{99}+dfrac{1}{100}1c) dfrac{1}{5}+dfrac{1}{6}+dfrac{1}{7}+...+dfrac{1}{17} 2d) dfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+...+dfrac{1}{29.30} 1
Đọc tiếp
\(\text{Bài 4. Chứng tỏ rằng:}\)
\(a\)) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}< 1\)
\(b\)) \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}>1\)
\(c\)) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
\(d\)) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}< 1\)