Tính
A=1-5^2+5^4+5^6+5^8-...+5^100
B= 1/3+(1/3)^2+(1/3)^3+...+(1/3)^2012
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Những câu hỏi liên quan
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 8 2023
A = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 +....+ 99 - 100
A = (1 - 2) + ( 3- 4) + ....+ (99 - 100)
Xét dãy số 1; 3;...; 99
Dãy số trên là dãy số cách đều với khoảng cách là: 3 - 1 = 2
Số số hạng của dãy số trên là: ( 99 - 1): 2 + 1 = 50
A là tổng của 50 nhóm mỗi nhóm cóa giá tri là: 1 - 2 = - 1
A = - 1 \(\times\) 50 = - 50
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
13 tháng 8 2023
B = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +...+ 97 - 98 - 99 + 100
B = ( 1 - 2 - 3 + 4) + ( 5 - 6 - 7 + 8) +...+ ( 97 - 98 - 99 + 100)
B = 0 + 0 +...+ 0
B = 0
6 tháng 4 2022
a.\(\dfrac{27}{8}\)
b.\(\dfrac{37}{40}\)
c.\(\dfrac{5}{2}\)
d.\(\dfrac{7}{3}\)
e.5
g.\(\dfrac{53}{16}\)
O
6 tháng 4 2022
Bài 1 :
a) \(\dfrac{3}{2}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{12}{8}+\dfrac{10}{8}+\dfrac{5}{8}=\dfrac{12+10+5}{8}=\dfrac{27}{8}\)
b) \(\dfrac{4}{5}-\dfrac{3}{8}+\dfrac{2}{4}=\dfrac{32}{40}-\dfrac{15}{40}+\dfrac{20}{40}=\dfrac{32-15+20}{40}=\dfrac{37}{40}\)
c) \(3+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{3}{1}+\dfrac{6}{8}-\dfrac{5}{4}=\dfrac{24}{8}+\dfrac{6}{8}-\dfrac{10}{8}=\dfrac{20}{8}=\dfrac{5}{2}\)
d) \(\dfrac{5}{6}-\dfrac{1}{2}+2=\dfrac{5}{6}-\dfrac{1}{2}+\dfrac{2}{1}=\dfrac{5}{6}-\dfrac{3}{6}+\dfrac{12}{6}=\dfrac{14}{6}=\dfrac{7}{3}\)
e) \(\dfrac{3}{5}+\dfrac{6}{11}+\dfrac{7}{13}+\dfrac{2}{5}+\dfrac{16}{11}+\dfrac{19}{13}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{6}{11}+\dfrac{16}{11}\right)+\left(\dfrac{7}{13}+\dfrac{19}{13}\right)=1+2+2=5\)
g) \(\dfrac{75}{100}+\dfrac{18}{21}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{3}{21}+\dfrac{13}{32}=\dfrac{3}{4}+\dfrac{6}{7}+\dfrac{29}{32}+\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{13}{32}=\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{6}{7}+\dfrac{1}{7}\right)+\left(\dfrac{29}{32}+\dfrac{13}{32}\right)=1+1+\dfrac{21}{16}=2+\dfrac{21}{16}=\dfrac{53}{16}\)
TT
2 tháng 5 2023
a)( \(\dfrac{-8}{40}\)+\(\dfrac{25}{40}\))+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-3}{8}\))
=\(\dfrac{17}{40}\)+(\(\dfrac{-15}{40}\))
=\(\dfrac{2}{40}\)
=\(\dfrac{1}{20}\)
b)\(\dfrac{-5}{21}\)-(\(\dfrac{16}{21}\)-\(\dfrac{21}{21}\))
=\(\dfrac{-5}{21}\)-\(-5\)
=\(\dfrac{-25}{105}\)-\(\dfrac{-105}{105}\)
=\(\dfrac{16}{21}\)
bằng đây thôi nha mỏi tay quá
12 tháng 5 2023
c: =3/7(-5/11+6/11)=3/7*1/11=3/77
d:=8/-3(4/3-7/3)=8/3
e: =3,4-5/8-0,4-3/8
=2-1
=1
14 tháng 4 2023
\(a,\dfrac{4}{7}+\dfrac{7}{2}\\ =\dfrac{8}{14}+\dfrac{49}{14}\\ =\dfrac{8+49}{14}\\ =\dfrac{57}{14}\)
\(b,\dfrac{5}{8}\times\dfrac{3}{2}\\ =\dfrac{5\times3}{8\times2}\\ =\dfrac{15}{16}\)
\(c,\dfrac{3}{2}\times\dfrac{5}{6}-\dfrac{2}{3}\\ =\dfrac{3\times5}{2\times6}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{2}{3}\\ =\dfrac{15}{12}-\dfrac{8}{12}\\ =\dfrac{15-8}{12}\\ =\dfrac{7}{12}\)
\(d,\dfrac{13}{15}+\dfrac{2}{5}:\dfrac{3}{4}\\ =\dfrac{13}{15}+\dfrac{2}{5}\times\dfrac{4}{3}\\ =\dfrac{13}{15}+\dfrac{2\times4}{5\times3}\\ =\dfrac{13}{15}+\dfrac{8}{15}\\ =\dfrac{13+8}{15}\\ =\dfrac{21}{15}\\ =\dfrac{7}{5}\)
KN
13 tháng 4 2023
a) \(\dfrac{-3}{20}\) + \(\dfrac{-7}{4}\) =\(\dfrac{-3}{20}\) + \(\dfrac{-35}{20}\) = -2
b) 6 và \(\dfrac{2}{3}\) - 4 và \(\dfrac{2}{3}\) = 2
c) \(\dfrac{-3}{10}\) + \(\dfrac{7}{12}\) = \(\dfrac{-18}{60}\) + \(\dfrac{35}{60}\) =\(\dfrac{17}{60}\)
d) \(\dfrac{35}{-9}\) . \(\dfrac{81}{7}\) = \(\dfrac{-35}{9}\) . \(\dfrac{81}{7}\) = 45
e) \(\dfrac{-2}{5}\) - \(\dfrac{-3}{4}\) = \(\dfrac{-8}{20}\) - \(\dfrac{-15}{20}\) = \(\dfrac{-8}{20}\) + \(\dfrac{15}{20}\) =\(\dfrac{7}{20}\)
f) \(\dfrac{5}{23}\) . \(\dfrac{7}{26}\) + \(\dfrac{5}{23}\) .\(\dfrac{9}{26}\) = \(\dfrac{5}{23}\) . ( \(\dfrac{7}{26}\) + \(\dfrac{9}{26}\) )= \(\dfrac{5}{23}\) . \(\dfrac{8}{13}\) = \(\dfrac{40}{299}\)
g) \(\dfrac{-3}{12}\) : \(\dfrac{4}{15}\) =\(\dfrac{-3}{12}\) . \(\dfrac{15}{4}\) =\(\dfrac{-5}{8}\)
h) 1 và \(\dfrac{1}{6}\) - 3 và \(\dfrac{1}{3}\) =\(\dfrac{7}{6}\) -\(\dfrac{10}{3}\) = \(\dfrac{-13}{6}\)
i) \(\dfrac{-2}{5}\) . (-3) + \(\dfrac{3}{8}\) . \(\dfrac{4}{-10}\) =(\(\dfrac{-2}{5}\) .\(\dfrac{-4}{10}\)) + [(-3) . \(\dfrac{3}{8}\)
= \(\dfrac{4}{25}\) + \(\dfrac{-9}{8}\) = \(\dfrac{32}{200}\) + \(\dfrac{-225}{200}\) = \(\dfrac{-193}{200}\)
j) \(\dfrac{-13}{17}\) + (\(\dfrac{13}{-21}\) + \(\dfrac{-4}{17}\) )
= ( \(\dfrac{-13}{17}\) + \(\dfrac{-4}{17}\) )+\(\dfrac{-13}{21}\)
= -1+\(\dfrac{-13}{21}\)
= \(\dfrac{-21}{21}\) + \(\dfrac{-13}{21}\) = \(\dfrac{-34}{21}\)
Khôi nguyễn
27 tháng 1 2016
Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh
5 tháng 4 2021
Bài 1:
Ta có: \(x-35\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow65\%\cdot x=\dfrac{1}{25}\)
\(\Leftrightarrow x=\dfrac{1}{25}:\dfrac{13}{20}=\dfrac{1}{25}\cdot\dfrac{20}{13}=\dfrac{4}{65}\)
Vậy: \(x=\dfrac{4}{65}\)
5 tháng 4 2021
Bài 2:
a) Ta có: \(17\dfrac{2}{31}-\left(\dfrac{15}{17}+6\dfrac{2}{31}\right)\)
\(=17\dfrac{2}{31}-\dfrac{15}{17}-6\dfrac{2}{31}\)
\(=11+\dfrac{2}{31}-\dfrac{15}{17}\)
\(=\dfrac{5366}{527}\)
21 tháng 12 2023
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
29 tháng 4 2022
a, \(1+\dfrac{3}{4}=\dfrac{7}{4}\)
b, \(\dfrac{4}{5}-\dfrac{3}{8}=\dfrac{32-15}{40}=\dfrac{17}{40}\)
c, \(1:\dfrac{2}{3}=\dfrac{1.3}{2}=\dfrac{3}{2}\)
d, \(\dfrac{2}{5}.\dfrac{5}{2}=1\)