\(A=\dfrac{7}{10}+\dfrac{7}{10^2}+...+\dfrac{7}{10^{2017}}+\dfrac{7}{10^{2018}}\)
\(10A=7+\dfrac{7}{10^1}+...+\dfrac{7}{10^{2016}}+\dfrac{7}{10^{2017}}\)
\(10A-A=7-\dfrac{7}{10^{2019}}\)
\(A=\dfrac{7.10^{2019}-7}{9.10^{2019}}\)
4, so sánh A và B:
a,A=\(\dfrac{3}{8^3}+\dfrac{7}{8^4}\);B=\(\dfrac{7}{8^3}+\dfrac{3}{8^4}\)
b,A=\(\dfrac{10^7+5}{10^7-8}\);B=\(\dfrac{10^8+6}{10^8-7}\)
c,A=\(\dfrac{10^{1992}+1}{10^{1991}+1}\);B=\(\dfrac{10^{1993}+1}{10^{1992}+1}\)
So sánh không qua quy đồng:\(A=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}};\) \(B=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}\)
So sánh giá trị 2 biểu thức:
A=\(\dfrac{1+7+7^2+...+7^9}{1+7+7^2+...+7^{10}}\) và B=\(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^{10}}\)
Mí bạn giúp tớ điii
giải rõ hộ nha :3
A =\(\dfrac{1+7^{ }+7^2+...+7^9}{1+7+7^2+...+7^{10}}\)
B =\(\dfrac{1+5+5^2+...+5^9}{1+5+5^2+...+5^{10}}\)
Hãy so sánh A & B
Bài 5 : So sánh
a) A= \(\dfrac{a+1}{a+2}\) và B= \(\dfrac{a+2}{a+3}\)
b) C= \(\dfrac{8^9+12}{8^7+7}\) và D =\(\dfrac{8^{10}+4}{8^{10}-1}\)
1. TÍNH NHANH
\(\dfrac{3}{7}+\dfrac{5}{3\cdot7}+\dfrac{7}{3\cdot19}+\dfrac{9}{7\cdot19}\)
2. KO QUY ĐỒNG HÃY SO SÁNH
\(\dfrac{3}{4};\dfrac{5}{6};\dfrac{7}{10}\)
a) \(\dfrac{\left(3+\dfrac{1}{6}\right)-\dfrac{2}{5}}{\left(5-\dfrac{1}{6}\right)+\dfrac{7}{10}}\)
b) \(\dfrac{\left(4,08-\dfrac{2}{25}\right):\dfrac{4}{17}}{\left(6\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{7}}\)
c) \(\dfrac{2-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{3}{5}}{3-\dfrac{1}{5}-\dfrac{5}{3}}\)
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
So sánh
A= \(\dfrac{7^{10}}{1+7+...+7^9}\) và
B=\(\dfrac{5^{10}}{1+5+...+5^9}\)
