Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ko đấy. đằng này tớ chưa học . help
tinh
1-2-3-4 - ........ - 99 +100
#)Giải :
a) \(A=\frac{4^5.9^4-2^6.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\frac{1}{3}\)
\(a,A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-1}{3}\)
Học tốt!!!!!!!!!!!!!
Tính
a)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100}\\ \)
\(=\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4.5...101}{2.3.4...100}\right)\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
b)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}\\ < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\\ \)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\\ =1-\frac{1}{n}< 1\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\left(\dfrac{2015}{2}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}}{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2015}+\dfrac{2017}{2016}}=\dfrac{1}{2017}\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}.....1\frac{1}{2015}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2016}{2015}\)
\(=\frac{3.4.5.....2016}{2.3.4....2015}=\frac{2016}{2}=1008\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2016}{2015}\)
\(A=\frac{2016}{2}=1008\)
Xong nhé bạn!
Ta có :
B = \(\frac{1^2+2^2+..................+2015^2}{2^2.1^2+2^2.2^2+.................+2^2.2015^2}\) = \(\frac{1^2+2^2+...........+2015^2}{2^2.\left(1^2+2^2+...............+2015^2\right)}\) = \(\frac{1}{2^2}=\frac{1}{4}\)
Vậy B = \(\frac{1}{4}\)
k nha bạn, mình nhanh nhất