Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a.\) \(\frac{6^3+3.6^2+3^3}{-13}=\frac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\frac{2^3.3^3+3^3.2^2+3^3}{-13}\)
\(=\frac{3^3.\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=\frac{3^3.\left(-1\right)}{1}=-27\)
\(b.\)\(A=2^2+4^2+6^2+...+20^2=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(A=2^2.\frac{10.\left(10+1\right).\left(2.10+1\right)}{6}=4.385=1540\)
( Ta có: công thức tính tổng bình phương liên tiếp tứ 1 đến n là: \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\))
\(c.\)\(B=100^2+200^2+...+1000^2=\left(100.1\right)^2+\left(100.2\right)^2+...+\left(100.10\right)^2\)
\(B=100^2.1^2+100^2.2^2+...+100^2.10^2=100^2.\left(1^2+2^2+...+10^2\right)\)
Áp dụng công thức \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
Ta có: \(B=100^2\times385=3,850,000\)
bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
Bài 1:
a) 510 : 57 - 25 . 30 = 53- 25= 125 - 32= 93
b) \(\frac{2^{14}\cdot14\cdot125}{35^3\cdot6}=\frac{2^{14}\cdot2\cdot7\cdot5^3}{\left(7\cdot5\right)^3\cdot2\cdot3}=\frac{2^{15}\cdot7\cdot5^3}{7^3\cdot5^3\cdot2\cdot3}=\frac{2^{14}}{7^2\cdot3}=\frac{16384}{49\cdot3}=\frac{16384}{174}\)
c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^{11}\cdot2^2+2^3\cdot2^2}{2^2\cdot2^8+2^2}=\frac{2^2\left(2^{11}+2^3\right)}{2^2\left(2^8+1\right)}=\frac{2^{11}+2^3}{2^8+1}=\frac{2^3\cdot2^8+2^3}{2^8+1}=\frac{2^3\left(2^8+1\right)}{2^8+1}=2^3=8\)
Bài 2:
a) 59 . 252 = 59 . (52)2= 59 . 54= 513
b) 1410 : 495 = [(2.7)10] : [(7.7)5] = (210 . 710 ) : (75 . 75) = 210 . 710 : 710= 210
c) 414 . 528 = (22)14 . 528 = 228 . 528 = (2.5)28 = 1028
d) 10010 : (-10)3 . (-100)4 = (102)3 : 103.(-1) . (102)4= 106 : 103 . 108 .(-1) = -1011
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
1) \(=\frac{6^5.5^3\left(1+5\right)}{6^5.5^3.3}=\frac{6}{3}=2\)
2)
\(2B=2+2^2+2^3+...+2^{101}\)
\(2B-B=B=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)=2^{101}-1\)