\(A=\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot...\cdot\dfrac{1}{2}=\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot...\cdot\dfrac{1}{2}=\dfrac{1}{100}\)
Tính hợp lí:
a) (100-9)(99-9)(98-9)...(1-9).
b) \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)\)
c) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\dfrac{99}{100}:\left(\dfrac{1}{4}-\dfrac{1}{12}+\dfrac{1}{3}\right)-\left(\dfrac{-7}{5}\right)^2\)
\(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
TÍnh: \(\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right).....\left(\dfrac{1}{100^2}-1\right)\)
Cho biểu thức \(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)
Hãy so sánh A với \(\dfrac{11}{19}\)
Tính giá trị biểu thức sau:
\(D=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(A=-5^{22}-\left\{-222-\left[-122-\left(100-5^{22}\right)+2022\right]\right\}\)
\(B=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+3+...+20\right)\)
\(C=\dfrac{5.4^6.9^4-3^9.\left(-8\right)^4}{4.2^{13}.3^8+2.8^4.\left(-27\right)^3}\)
\(\left(\dfrac{1}{2^2}-1\right)\).\(\left(\dfrac{1}{3^2}-1\right).\left(\dfrac{1}{4^2}-1\right)\).....\(\left(\dfrac{1}{100^2}-1\right)\)
giúp mình nhanh với. tối mình phải nộp rùi huhuhu
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{99}\right)\)
B= (\(1+\dfrac{1}{2}\))\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)...\left(1+\dfrac{1}{99}\right)\)