thực hiện phép tính.
a) b) C= 3+31+32+33+...+3100 c)\(\dfrac{2018.2019-1}{2018^2+2017}\)
\(\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
K
Khách
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KV
31 tháng 10 2023
a) 2021 - (1/3)² . 3²
= 2021 - 1/9 . 9
= 2021 - 1
= 2020
b) 5/10 + 9 . (-3/2)
= 1/2 - 27/2
= -26/2
= -13
c) -10 . (-2021/2022)⁰ + (2/5)² : 2
= -10 . 1 + 4/25 . 2
= -10 + 8/25
= -68/7
DD
31 tháng 10 2023
\(a,2021-\left(\dfrac{1}{3}\right)^2\cdot3^2\\ =2021-\dfrac{1}{9}\cdot9\\ =2021-\dfrac{9}{9}\\ =2021-1=2020\\ b,\dfrac{5}{10}+9\cdot\dfrac{-3}{2}\\ =\dfrac{5}{10}+\dfrac{-27}{2}\\ =\dfrac{5}{10}+\dfrac{-135}{10}\\ =-\dfrac{130}{10}\\ =-13\\ c,-10\cdot\left(-\dfrac{2021}{2022}\right)^0+\left(\dfrac{2}{5}\right)^2:2\\ =-10\cdot1+\dfrac{4}{25}\cdot\dfrac{1}{2}\\ =-10+\dfrac{4}{50}\\ =-10+\dfrac{2}{25}\\ =-\dfrac{248}{25}\)
5 tháng 7 2018
\(A=\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{3^{10}\cdot\left(11+5\right)}{3^9\cdot16}=\dfrac{3^{10}\cdot16}{3^9\cdot16}=3\)
\(B=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\dfrac{2^{10}\cdot\left(13+65\right)}{2^8\cdot2^2\cdot26}=\dfrac{2^{10}\cdot78}{2^{10}\cdot26}=3\)
\(C=\dfrac{72^3\cdot54^2}{108^4}=\dfrac{\left(2^3\cdot3^2\right)^3\cdot\left(2\cdot3^3\right)^2}{\left(3^3\cdot2^2\right)^4}\\ =\dfrac{2^9\cdot3^6\cdot2^4\cdot3^6}{3^{12}\cdot2^8}=\dfrac{2^{13}\cdot3^{12}}{3^{12}\cdot2^8}=2^5=32\)
\(D=\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\dfrac{11\cdot3^{29}-\left(3^2\right)^{15}}{2^2\cdot3^{28}}=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}\\ =\dfrac{3^{29}\cdot\left(11-3\right)}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{4\cdot3^{28}}=3\cdot2=6\)
26 tháng 2 2022
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
FC
25 tháng 10 2018
a, \(B=\dfrac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(=\dfrac{2^{10}.\left(13+65\right)}{2^8.2^3.13}\)
\(=\dfrac{2^{10}.78}{2^{11}.13}\)\(=\dfrac{1.6}{2.1}=\dfrac{1.3}{1.1}=3\)
29 tháng 10 2022
b: \(=\dfrac{2^{20}\cdot3^2+2^{54}}{2^{18}\cdot5^2}=\dfrac{2^{20}\left(3^2+2^{32}\right)}{2^{18}\cdot5^2}=\dfrac{2^2\left(3^2+2^{32}\right)}{25}\)
c: \(=\dfrac{2^9\cdot3^6\cdot3^6\cdot2^2}{2^8\cdot3^{12}}=\dfrac{2^{11}}{2^8}=8\)
d: \(=\dfrac{2^{12}\cdot3^4\cdot3^{10}}{2^{12}\cdot3^{12}}=9\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
10 tháng 9 2023
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
KV
10 tháng 9 2023
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
2 tháng 12 2018
A=\(\dfrac{2^{10}\left(13+65\right)}{2^8.104}\)
=\(\dfrac{2.78}{104}\)=\(\dfrac{78}{52}\)=\(\dfrac{39}{26}\)
17 tháng 9 2023
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
Sửa đề: \(C=1+3^1+3^2+...+3^{100}\)
b) Ta có: \(C=1+3^1+3^2+...+3^{100}\)
\(\Leftrightarrow3\cdot C=3+3^2+...+3^{101}\)
\(\Leftrightarrow C-3\cdot C=1+3+3^2+...+3^{100}-3-3^2-...-3^{100}-3^{101}\)
\(\Leftrightarrow-2\cdot C=1-3^{101}\)
hay \(C=\dfrac{3^{101}-1}{2}\)
b) Ta có: C=1+31+32+...+3100C=1+31+32+...+3100
⇔3⋅C=3+32+...+3101⇔3⋅C=3+32+...+3101
⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101
⇔−2⋅C=1−3101