So sánh A ;B : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2};B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
Cho A= \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) và B =\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\). Khi đó \(\frac{A}{B}\)= ?
\(ChoA=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}vàB=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\).
Khi đó \(\frac{A}{B}=?\)
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)và \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\). Vậy \(\frac{A}{B}\)=...
Cho A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) và B=\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\). Khi đó \(\frac{A}{B}\)=
Cho \(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
so sánh B với \(\frac{3}{4}\)
Cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}\)và \(B=\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{200^2}\)
Khi đó \(\frac{A}{B}=.....\)
cho \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
và \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
Khi đó \(\frac{A}{B}=?\)
1. So sánh:
a. 1 và \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{50}}\)
b. \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^{100}}\)với \(\frac{1}{2}\)
c. \(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^6}+.....\frac{1}{4^{1000}}\)với \(\frac{1}{3}\)
2. Tìm x, biết:
a.\(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
b.\(3-\frac{1-\frac{1}{2}}{1+\frac{1}{x}}=2\frac{2}{3}\)
c.\(4^x+4^{x+3}=4160\)
d.\(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
e.\(\frac{x-100}{24}+\frac{x-98}{26}+\frac{x-96}{24}=3\)
g.\(\frac{x-1}{65}+\frac{x-3}{63}+=\frac{x-5}{61}+\frac{x-7}{59}\)