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19 tháng 12 2019
\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{49}+\left(-5\right)^{50}\\ -5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+\left(-5\right)^4+...+\left(-5\right)^{50}+\left(-5\right)^{51}\\ B+5B=\left[\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{49}+\left(-5\right)^{50}\right]-\left[\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+\left(-5\right)^4+...+\left(-5\right)^{50}+\left(-5\right)^{51}\right]\\ 6B=\left(-5\right)^0-\left(-5\right)^{51}\\ B=\frac{1-\left(-5\right)^{51}}{6}\)
19 tháng 5 2017
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{43}}\)
\(A=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}\)
\(A=\frac{5^{30}.7^{48}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}=5.7^3.\left(1-7.2^2\right)=1715.\left(-27\right)=-46305\)
E
19 tháng 5 2017
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}\left(2^4\right)^2.7^{43}}=\frac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{43}}=\frac{7^{48}.5^{30}.2^8\left(1-7.2^2\right)}{5^{29}.2^8.7^{43}}\)
=\(7^5.5.\left(-27\right)=-2268945\)
BD
2
BB
7 tháng 2 2020
-5B=(-5)1+(-5)2+(-5)3+...+(-5)2018
-5B-B=[(-5)1+(-5)2+...+(-5)2018] - [(-5)0+(-5)1+...+(-5)2017]
-6B=(-5)2018-(-5)0 = (-5)2018-1
B= [(-5)2018-1]:-6
Anh học tốt nha ( em mới lớp 6)
3 tháng 2 2022
\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8-5^{30}\cdot7^{49}\cdot2^{10}}{3^2\cdot2^2\cdot5^2\cdot7^{48}}\)
\(=\dfrac{7^{48}\cdot5^{30}\cdot2^8\left(1+7\cdot4\right)}{3^2\cdot2^2\cdot5^2\cdot7^{48}}=\dfrac{5^{28}\cdot2^6\cdot12}{3^2}=\dfrac{5^{28}\cdot2^8}{3}\)
NH
0
28 tháng 11 2017
\(A=\frac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
\(A=\frac{\left(7^2\right)^{24}.\left(5^3\right)^{10}.2^8-5^{30}.7^{49}.\left(2^2\right)^5}{5^{29}.\left(2^4\right)^2.7^{48}}\)
\(A=\frac{7^{49}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
\(A=\frac{7^{48}.5^{30}.2^8\left(1-28\right)}{5^{29}.2^8.7^{48}}\)
\(A=5.\left(-27\right)\)
\(A=-135\)
HM
2
27 tháng 2 2017
Ta có : A = 1 + 5 + 52 + ...... + 549 + 550
=> 5A = 5 + 52 + 53+..... + 550 + 551
=> 5A - A = 551 - 1
=> 4A = 551 - 1
=> \(A=\frac{5^{51}-1}{4}\)
28 tháng 2 2017
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
5A=\(5+5^2+5^3+...+5^{50}+5^{51}.\)
5A-A=\(\left(5+5^2+5^3+.....+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}.\right)\)
4A=\(5^{51}-1\)
\(=>A=\frac{5^{51}-1}{4}\)
LB
30 tháng 6 2017
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
\(5A=5^1+5^2+5^3+5^4+...+5^{51}\)
\(4A=5A-A=5^{51}-1\)
\(\Rightarrow A=\frac{5^{51}-1}{4}\)
b/
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
\(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{100}\)
\(\frac{1}{2}B=B-\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\)
\(B=\frac{1}{2}B\cdot2=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)
\(B=1-\frac{1}{2^{99}}\)
TT
1
2 tháng 10 2017
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\)
=> \(5\text{A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\)
=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - (\(1+5+5^2+5^3+...+5^{49}+5^{50}\) )
=> \(5\text{A-A}=5+5^2+5^3+5^4...+5^{49}+5^{50}+5^{51}\) - \(1-5-5^2-5^3-...-5^{49}-5^{50}\)
=> \(4\text{A}=5^{51}-1\)
=> \(A=\dfrac{5^{51}-1}{4}\)
Rút gọn:
\(A=5^0+5^1+5^2+...+5^{99}+5^{50}\)
\(5A=5^1+5^2+5^3+...+5^{51}\)
\(5A-A=\left(5^1+5^2+5^3+...+5^{51}\right)-\left(5^0+5^1+5^2+...+5^{50}\right)\)
\(4A=5^{51}-5^0\)
\(=>A=\left(5^{51}-5^0\right):4\)
Vậy : \(A=\left(5^{51}-5^0\right):4\)