QG
5
K
Khách
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NA
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 9 2023
A = 1 + 3 + 5 + 7 +...+ 101
Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1 = 2
Số số hạng của dãy số trên là: (101 - 1):2 +1 = 51
Tổng dãy số trên là: A = (101 + 1) x 51: 2 = 2601
Q
2 tháng 4 2017
Bài 1: Tính
\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)
\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)
\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)
\(=\dfrac{5}{8}.\dfrac{-4}{15}\)
\(=\dfrac{-1}{6}\)
\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{3}{4}\)
\(=\dfrac{-51}{40}\)
\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)
\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)
\(=\dfrac{4}{6}\)
\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)
\(=\dfrac{4}{3}-1\)
\(=\dfrac{1}{3}\)
\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)
\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)
\(=1:\dfrac{1}{5}\)
\(=5\)
\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)
\(=3+\dfrac{11}{7}\)
\(=3\dfrac{11}{7}=\dfrac{32}{7}\)
\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)
\(=1:\dfrac{19}{5}\)
\(=\dfrac{5}{19}\)
\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)
\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+1\right)\)
\(=1:\dfrac{5}{3}\)
\(=\dfrac{3}{5}\)
\(\text{9)}\)
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)
\(=\dfrac{330875}{1507764}\)
2 tháng 4 2017
1) \(\frac{5}{8}.\frac{7}{3}-\frac{5}{2}.\frac{1}{8}=\frac{5}{8}.\frac{7}{3}-\frac{5}{8}.\frac{1}{2}=\frac{5}{8}\left(\frac{7}{3}-\frac{1}{2}\right)=\frac{5}{8}.\frac{11}{6}=\frac{55}{48}\)
2) \(\frac{21}{10}.\frac{3}{4}-\frac{21}{10}.\frac{3}{4}=\frac{21}{10}\left(\frac{3}{4}-\frac{3}{4}\right)=\frac{21}{10}.0=0\)
3) \(\frac{-4}{11}:\frac{-6}{11}=\frac{-4}{11}.\frac{-11}{6}=\frac{-4.\left(-11\right)}{11.6}=\frac{-4.\left(-1\right)}{1.6}=\frac{4}{6}=\frac{2}{3}\)
4)\(\frac{2}{7}.\frac{14}{3}-1=\frac{2.14}{7.3}-1=\frac{2.2}{1.3}-1=\frac{4}{3}-1=\frac{1}{3}\)
5)\(\frac{4}{7}:\left(\frac{1}{5}.\frac{4}{7}\right)=\frac{4}{7}:\frac{4}{35}=\frac{4}{7}.\frac{35}{4}=\frac{4.35}{7.4}=\frac{1.5}{1.1}=5\)
6) \(\frac{12}{7}.\frac{7}{4}+\frac{35}{11}:\frac{245}{121}=\frac{12.7}{7.4}+\frac{35}{11}.\frac{121}{245}=\frac{3.1}{1.1}+\frac{35}{11}.\frac{121}{245}=3+\frac{35}{11}.\frac{121}{245}=3+\frac{35.121}{11.245}=\frac{1.11}{1.7}=\frac{11}{7}\)
LT
1
13 tháng 10 2016
S = 1+2-3-4+5+6-7-8+...............+193+194-195-196+197+198
S = 1+(2-3-4+5)+(6-7-8+9) +.....+(194-195-196+197) +198
S = 1+ 0 + 0 +......+ 0 +198
S = 1+198
S =199
quá dễ
25 tháng 6 2019
P = \(\frac{1}{3}\)+ \(\frac{13}{15}\)+\(\frac{33}{35}\)+...+\(\frac{193}{195}\)(7 số hạng)
P = (\(\frac{3}{3}\)-\(\frac{2}{3}\)) + ( \(\frac{15}{15}\)-\(\frac{2}{15}\) ) + ... + (\(\frac{195}{195}\)-\(\frac{2}{195}\) )
P = (1+1+1+..+1) -(\(\frac{2}{3}\)+\(\frac{2}{15}\)+\(\frac{2}{35}\)+...+\(\frac{2}{195}\))
P = 7 - (\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+...+\(\frac{2}{13.15}\))
P = 7 - (1 - \(\frac{1}{3}\)+ \(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{13}\)-\(\frac{1}{15}\))
P = 7 - (1 - \(\frac{1}{15}\))
P = 7 - 1 +\(\frac{1}{15}\)
P = 6 + \(\frac{1}{15}\)
=> P > 6
LN
10 tháng 5 2020
Bài 1: Tính hợp lý (nếu có thể)
a) (-193)+36+14+193
=[(-193)+193]+(36+14)
=50
b) 2008-(127+2008)+(-35+127)
=2008-127-2008-35+127
=(2008-2008)+(127-127)-35
=-35
c) (273-28)+(129-72)
=273-28+129-72
=302
d) 21×35-5×11×7
=21.35-11.35
=(21-11).35
=10.35=350
e) (-13)×34-87×34
=34(-13-87)
=34.(-100)
=-3400
f) 85×(35-27)-35×(85-27)
=85.35-85.27-35.85+35.27
=(85.35-35.85)+27(-85+35)
=0+27.(-50)=-1350
g) 1-2-3+4+5-6-7+…+97-98-99+100
=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100) *
Dãy trên có 100 số hạng chia thành 25 nhóm mỗi nhóm 4 số hạng mối nhóm đều có kết quả bằng 0
*=0+0+..+0( 25 số hạng)=0
h) A=2100-299-298-....-22-2-1
2A=2101-2100-299-...-23-22-2
⇒2A-A=(2101-2100-299-...-23-22-2)-(2100-299-298-....-22-2-1)
⇒A=2101-2100-299-...-23-22-2-2100+299+298+....+22+2+1
⇒A=2101-2.2100+1
⇒A=2101-2101+1
⇒A=0+1=1
Ý bạn là sao?= 193 à
1+3+5+7+...+193
= (193+1).97 : 2
= 194.97 .1/2
= 97.(194.1/2)
= 97.97
= 9409