THỰC HIỆN PHÉP TÍNH SAU:
a/\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...........+\(\dfrac{1}{99.100}\)
b/\(\dfrac{5.4^{15}-9^9-4.3^{10}.8^9}{5.29-6^{19}.7.2^{29}.27^6}\)
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Những câu hỏi liên quan
MN
1
16 tháng 6 2023
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3-7\right)}=\dfrac{10-9}{15-7}=\dfrac{1}{8}\)
31 tháng 12 2022
a)= 2021.2021-2020.(2021+1)
= 2021.(2020+1)-2020.(2021+1)
= (2021.2020)+2021-(2020.2021)-2020
= 1
31 tháng 12 2022
b) B= (1+2-3-4)+(5+6-7-8)+(9+10-11-12)...........+(2017+2018-2019-2020)+2021
B= -4+(-4)+....................(-4)+2021
B= -4x505+2021
B= -2020 + 2021
B = 1
8 tháng 10 2023
a: \(A=\dfrac{3^3\cdot2^3+3^3\cdot2^2+3^3\cdot1}{-13}=\dfrac{27\left(2^3+2^2+1\right)}{-13}=-27\)
b: \(B=\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^3\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9}{2^{10}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\right)}{2^{10}\cdot3^7\left(1+5\cdot3\right)}=\dfrac{2}{3}\cdot\dfrac{4+27}{1+15}=\dfrac{2}{3}\cdot\dfrac{31}{16}=\dfrac{31}{24}\)
c: \(C=\dfrac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{35}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot2^6-7\right)}=\dfrac{10-9}{5\cdot64-7}=\dfrac{1}{313}\)
LL
1
ND
Nguyễn Đức Trí
VIP
27 tháng 8 2023
\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\) (sửa \(-9^9\rightarrow.9^9\))
\(=\dfrac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{2^{29}.3^{18}\left(15-9\right)}{2^{28}.3^{18}\left(15-14\right)}\)
\(=2.6=12\)
TL
1
SD
1
TD
21 tháng 3 2017
\(\dfrac{5.4^{15}.9^9+4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}=\dfrac{5.2^{30}.3^{18}+2^2.3^{20}.2^{27}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
=\(\dfrac{5.2^{30}.3^{18}+2^{29}.3^{20}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}=\dfrac{2^{29}.3^{18}\left(5.2+3^2\right)}{2^{29}.3^{18}\left(5.3-7\right)}\)
=\(\dfrac{10+9}{15-7}=\dfrac{19}{8}\)
NT
1
MC
27 tháng 6 2018
\(\dfrac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\dfrac{5.3^{20}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\dfrac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{28}.3^{18}\left(5.3-7.2\right)}\)
\(=2\)
a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)
=\(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
a) 11.2+12.3+13.4+....+199.10011.2+12.3+13.4+....+199.100
= 11−12+12−13+13−14+....+199−110011−12+12−13+13−14+....+199−1100
=11+0+0+...+0−110011+0+0+...+0−1100
=1−11001−1100
= 99100