\(\text{Ta có:}8^4>8^3\Rightarrow\frac{4}{8^4}< \frac{4}{8^3}\Rightarrow\frac{3}{8^3}+\frac{3}{8^4}+\frac{4}{8^4}< \frac{3}{8^3}+\frac{4}{8^3}+\frac{3}{8^4}\Rightarrow A< B\)
Các câu hỏi tương tự
So sánh
\(A=40+\frac{3}{8}+\frac{7}{8^2}+\frac{5}{8^3}+\frac{32}{8^5}\)
\(B=\frac{24}{8^2}+40+\frac{5}{8^2}+\frac{40}{8^4}+\frac{5}{8^4}\)
Cho:
\(A=40+\frac{3}{8}+\frac{7}{8^2}+\frac{5}{8^3}+\frac{32}{8^5}\)
\(B=\frac{24}{8^2}+40+\frac{5}{8^2}+\frac{40}{8^4}+\frac{5}{8^4}\)
Hãy so sánh A và B
cho A =\(40+\frac{3}{8}+\frac{7}{8^2}+\frac{5}{8^3}+\frac{32}{8^5}\)
B=\(\frac{24}{8^2}+40+\frac{5}{8^2}+\frac{40}{8^4}+\frac{5}{8^4}\)
So sánh A và B
Cho \(A=\frac{3}{8^3}+\frac{-7}{8^4}\)và \(B=\frac{-7}{8^3}+\frac{3}{8^4}\)
Hãy so sánh A và B
So sánh:
\(A=\frac{3}{8^3}+\frac{7}{8^4}\) và \(B=\frac{7}{8^3}+\frac{3}{8^4}\)
giải chi tiết hộ mình nha ai nhanh nhất **** cho
So sánh \(D=\frac{3}{8^3}+\frac{7}{8^4}\)và \(C=\frac{3}{8^4}+\frac{7}{8^3}\)
1)So sánh:
A=\(\frac{3}{8^3}\) +\(\frac{7}{8^4}\) và B=\(\frac{7}{8^3}\)+\(\frac{3}{8^4}\)
Tính:
a) \(A=\frac{\frac{7}{8}+\frac{7}{27}-\frac{7}{49}}{\frac{11}{8}+\frac{11}{27}-\frac{11}{49}}\)
b)\(B=\frac{\frac{8}{9}-\frac{8}{27}-\frac{8}{81}+\frac{8}{243}}{4-\frac{4}{3}-\frac{4}{9}+\frac{4}{27}}\)
c)\(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
a) so sánh \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+\frac{5}{6}\) và 4
b)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)và 1
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)\(A=\frac{2011}{2012}+\frac{2012}{2013};B=\frac{2011+2012}{2012+2013}\)