Các câu hỏi tương tự
so sanh A,B .A=\(\frac{10^9+1}{10^8+1}\) ,,B\(\frac{10^{20}+1}{10^{21}+1}\)
So sanh A va B, biet :
a)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
b)\(A=\frac{7^{10}}{1+7+7^2+...+7^9};B=\frac{5^{10}}{1+5+5^2+...+5^9}\)
cho A=10^7+1/10^8+1
B=10^8+1/10^9+1
so sanh A va B
So sánh:
a, A= \(\frac{10^8+2}{10^8-1}\) ; B= \(\frac{10^8}{10^8-3}\)
b, A= \(\frac{8^{10}+1}{8^{10}-1}\) ; B=\(\frac{8^{10}-1}{8^{10}-3}\)
c, A= \(\frac{100^9+4}{100^9-1}\): B= \(\frac{100^9+1}{100^9-4}\)
bài 1 So sánha)Afrac{3}{8^3}+frac{7}{8^4} ; Bfrac{7}{8^3}+frac{3}{8^4}b)Afrac{10^{1992}+1}{10^{1991}+1};Bfrac{10^{1993}+1}{10^{1992}+1}c)Afrac{10^7+5}{10^4-8};Bfrac{10^8+6}{10^8-7}d)Afrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};Bfrac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}e)Afrac{2011}{2012}+frac{2012}{2013};Bfrac{2011+2012}{2012+2013}
Đọc tiếp
bài 1 So sánh
a)\(A=\frac{3}{8^3}+\frac{7}{8^4}\) ; \(B=\frac{7}{8^3}+\frac{3}{8^4}\)
b)\(A=\frac{10^{1992}+1}{10^{1991}+1};B=\frac{10^{1993}+1}{10^{1992}+1}\)
c)\(A=\frac{10^7+5}{10^4-8};B=\frac{10^8+6}{10^8-7}\)
d)\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8};B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}\)
e)\(A=\frac{2011}{2012}+\frac{2012}{2013};B=\frac{2011+2012}{2012+2013}\)
so sanh 2 PS sau
\(\frac{7^{10}+1}{7^{10}-1}\)va \(\frac{7^{10}-1}{7^{10}-3}\)
Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
So sánh:
\(A=\frac{10^8+1}{10^9+1}\)với\(B=\frac{10^9+1}{10^{10}+1}\)
So sanh cac phan so sau:
A=\(\frac{3^{10}+1}{3^9+1}\)va B=\(\frac{3^9+1}{3^8+1}\)