Mình làm được một câu thôi, bạn dựa vào làm nha!
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{98^2}+\dfrac{1}{99^2}+\dfrac{1}{100^2}< 1\)
Tính hợp lí:
A=\(\dfrac{3}{1.3}+\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{49.51}\)
B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^8}\)
Giúp mik nha mik đang cần rất là gấp nha !!!!!!!!!!
A=\(\dfrac{2}{3}\)+\(\dfrac{14}{15}\)+\(\dfrac{34}{35}\)+\(\dfrac{62}{63}\)+\(\dfrac{98}{99}\)+\(\dfrac{142}{143}\)+\(\dfrac{194}{195}\)
Và B=5+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^3}\)+\(^{\dfrac{1}{4^4}}\)+\(\dfrac{1}{5^5}\)+\(\dfrac{1}{6^6}\)+\(\dfrac{1}{7^7}\).So sánh A và B
Tính giá trị của biểu thức:
\(A=\dfrac{1}{2}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2^{100}}\)
Nhanh nhé mình cần gấp lắm!!!
\(\dfrac{1}{7^2}+\dfrac{1}{7^3}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{99}}+\dfrac{1}{7^{100}}\)
B=(1-\(\dfrac{1}{2^2}\))(1-\(\dfrac{1}{3^2}\))(1-\(\dfrac{1}{4^2}\))(1-\(\dfrac{1}{5^2}\))...(1-\(\dfrac{1}{99^2}\))(1-\(\dfrac{1}{100^2}\))
C=(\(\dfrac{1}{4}\)-1)(\(\dfrac{1}{9}\)-1)(\(\dfrac{1}{16}\)-1)...(\(\dfrac{1}{999}\)-1)
\(\dfrac{99}{100}:\left(\dfrac{1}{4}-\dfrac{1}{12}+\dfrac{1}{3}\right)-\left(\dfrac{-7}{5}\right)^2\)
\(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
