1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1, so sánh A;B biết: A=\(\left(\dfrac{\left(3\cdot\dfrac{2}{15}+\dfrac{1}{5}\right):2\cdot\dfrac{1}{2}}{\left(5\cdot\dfrac{3}{7}-2\cdot\dfrac{1}{4}\right):\dfrac{443}{56}}\right);B=\dfrac{1,2:\left(1\cdot\dfrac{1}{5}.1\cdot\dfrac{1}{4}\right)}{0,32+\dfrac{2}{25}}\)
1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
1: Tính B
\(B=1+\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{100}\cdot\left(1+2+3+...+100\right)\)
tinh
(1-\(\dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\)
1: \(\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{2017}\cdot\left(1+2+3+...+2017\right)\)
2: tính hợp li
a, \(44\cdot82-20^2+18\cdot44\)
b, \(\left(6^{10}:6^8\right):\left\{780:\left[78\cdot5-\left(125\cdot7^2\right)+13\cdot5\right]\right\}\)
Tính các tích sau:
a) \(P=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
b) \(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)
\(4\cdot\left(\dfrac{1}{2}-x\right)-5\cdot\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
Tính :
a, \(\dfrac{3\cdot13-13\cdot18}{15\cdot40-80}\);
b, \(\dfrac{18\cdot34+\left(-18\right)\cdot124}{-36\cdot17+9\cdot\left(-52\right)}\);
c, \(\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}+\dfrac{-0,25\cdot\dfrac{-2}{3}-0,75:\left(\dfrac{-1}{2}+\dfrac{2}{3}\right)}{\left|-1\dfrac{1}{2}\right|\cdot\left(\dfrac{-2}{3}-75\%:\dfrac{3}{-2}\right)}\).