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Câu 1:120a chia hết cho 12
36b chia hết cho 12
=>(120a+36b)chia het cho 12
\(Cau2:\)\(5^7+5^6+5^5\)=\(5^5\left(5^2+5+1\right)=5^5\cdot21\)
=>\(5^7+5^6+5^5chiahetcho21\)
C1: Vì 120 chia hết cho 12 nên 120a chia hết cho 12. (1)
Vì 36 chia hết cho 12 nên 36b chia hết cho 12. (2)
Từ (1) và (2) => (120a + 36b) chia hết cho 12
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a) \(4^3\cdot32^4\)
\(=\left(2^2\right)^3\cdot\left(2^5\right)^4\)
\(=2^6\cdot2^{20}\)
\(=2^{26}\)
b) \(3^{20}\cdot9^{10}\cdot27^2\)
\(=3^{20}\cdot\left(3^2\right)^{10}\cdot\left(3^3\right)^2\)
\(=3^{20}\cdot3^{20}\cdot3^6\)
\(=3^{46}\)
c) \(3^{10}\cdot7^{10}\)
\(=\left(3\cdot7\right)^{10}\)
\(=21^{10}\)
d) \(6^{15}:6^{14}\)
\(=6^{15-14}\)
\(=6\)
e) \(28^3:7^3\)
\(=4^3\cdot7^3:7^3\)
\(=4^3\)
\(=2^6\)
a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
\(a,\)\(\frac{2^5\times3^{12}\times7^8}{2^7\times3^{10}\times7^9}=\frac{3^2\times\left(2^5\times3^{10}\times7^8\right)}{2^2\times7\times\left(2^5\times3^{10}\times7^8\right)}\)\(=\frac{3^2}{2^2\times7}=\frac{9}{28}\)
\(b,\)Tương tự
a,\(\dfrac{3^6.5^7.7^{11}}{3^4.5^7.7^{10}}=\dfrac{3^4.3^2.5^7.7^{10}.7}{3^4.5^7.7^{10}}\) \(=9.7=63\)
b,\(\dfrac{2^{43}+2^4}{2^{39}+1}=\dfrac{2^{39}.2^4+2^4}{2^{39}+1}\) \(=\dfrac{2^4\left(2^{39}+1\right)}{2^{39}+1}=16\)
tính tổng hả bạn
vâng bạn
A = 71 + 72 + 73 + 74 + ...............................+ 7 2019
= 7 ( 1+2+3+4+..............+2019)
=> 1+2+3+4...........+2019)
= (1+2019) x 2019 :2
= 2039190
=> A = 7(2039190)
câu b
B = 101 + 102 +103 + ................................+102019
B = 10 (1+2+3+...................+2019)
=> (1+2+3+................+2019)
= (1+2019) x 2019 : 2
= 2039190
=> B = 10 2039190
Tần Nguyễn Phương Vy sai rồi. Mất căn bản nhá.
101+102 = 110
mà 10(1+2) lại bằng 1000
Sai hết rồi nhá
Giải:
\(A=7^1+7^2+7^3+7^4+...+7^{2019}\)
\(7A=7^1.7+7^2.7+7^3.7+7^4.7+...+7^{2019}.7\)
\(7A=7^2+7^3+7^4+...+7^{2019}+7^{2020}\)
\(7A-A=6A=\left(7^2+7^3+7^4+...+7^{2019}+7^{2020}\right)-\left(7^1+7^2+7^3+...+7^{2018}+7^{2019}\right)\)
\(6A=7^2+7^3+7^4+...+7^{2019}+7^{2020}-7^1-7^2-7^3-...-7^{2018}-7^{2019}\)
\(6A=-7^1+7^2-7^2+7^3-7^3+7^4-7^4+......+7^{2019}-7^{2019}+7^{2020}\)
\(6A=7^{2020}-7\)
\(A=\frac{7^{2020}-7}{6}\)
Câu b tương tự