c/m 1/2mu 3 +1/3 mu 3 +..................+1/ 2006 mu 3<1/4
c/m 1/2 mu 3 +1/3 mu 3 +..................+1/2006 mu3 >1/15
chung to rang B = 1/2mu 2 cong 1/3 mu 2 cong 1/4 mu 2 cong 1/5 mu 2 cong 1/6 mu 2cong 1/7 mu 2 cong 1/8 mu2 nho hon 1
so chinh phuong la so bang binh phuong cua mot so tu nhien ( vi du 0 ,1,2,4,9,16,...0) moi tong sau day co so chinh phuong ko
a) 1mu 3 +2 mu 3
b)1mu 3+2mu 3 +3 mu 3
c)1 mu 3+ 2 mu 3 + 3 mu 3 +4 mu 3
tai minh ko biet go dau mu !
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016 < 0,2
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016<0,2
Cm 1/2 mu 2 - 1/ 2mu 4 + 1/ 2 mu 6-...-1/2mu 4n -2 -1/2 mu 4n + ...+ 1/ 2 mu 2014 - 1/ 2 mu 2016<0,2
\(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2014}}-\frac{1}{2^{2016}}\)
\(\Rightarrow2^2A=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-...+\frac{1}{2^{2012}}-\frac{1}{2^{2014}}\)
\(\Rightarrow2^2A+A=1+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{2014}}-\frac{1}{2^{2014}}\right)-\frac{1}{2^{2016}}\)
\(\Rightarrow5A=1-\frac{1}{2^{2016}}< 1\Rightarrow A< \frac{1}{5}=0,2\)
đây là toán lớp 2 hả?
đây là toán lớp mấy thế
chứng minh ý a 8mu10 cong 2 mũ 20 chia hết cho41
chứng minh ý b 31 mu36 nhận 36 -313 mu 5 nhận 299 chia hết ch
chung minhý c 3 mu n công 3 cộng 2 mu n công 3 công 2 mu n công 2 công 2 mu n công 2 chia hết 6
chung minh y d d=2 cong 2 mu 2 cong 2mu 3 cong 2 mu 4 cong 2 mu 5 cong 2 mu 6 cong ....cong 2 mu 58 cong 2 mu 59 cong 2 mu 60 chia het cho 31
chung minh y e, e= 1cong 3 cong 3 mu 2 cong 3 mu 3 cong..cong 3 mu 98 cong 3 mu 99 chung minh e chia het cho5 chia het cho11
CHO A= 3+3MU2+3mu3+3mu4+...+3mu2017 a) tim so tu nhien N biet 2A +3 = 3n b)tim chu so tan cung cua A
bai1
5mu3+3mu5
bai2
(x-1)mu3=125
720/(41-(2*x-5))=2mu 3*5
bai3
1 phan 9 * 3 mu 4 * 3 mu n =3 mu 7
(2 mu 2 chia 4)* 2 mu n = 4
bai4
2 * 2mu2 * 2mu3 * 2mu4 *.......*2 mu100
A=1/2 +1/2mu 2+...+1/2 mu 15
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{15}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{14}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{15}}\right)\)
\(A=1-\frac{1}{2^{15}}\)\
\(A=1-\frac{1}{32768}\)
\(A=\frac{32767}{32768}\)
tim so tu nhien n,biet: a) (2mu n+1)=27 b) (n-2)mu 2=(n-2)mu 4 c)3 mu n+1=81 d)2mu2n+1=512