A= \(\left(\dfrac{1}{2}-1\right)\)\(\left(\dfrac{1}{3}-1\right)\).........\(\left(\dfrac{1}{10}-1\right)\). So sánh A với \(\dfrac{-1}{9}\)
B= \(\left(\dfrac{1}{4}-1\right)\)\(\left(\dfrac{1}{9}-1\right)\)...........\(\left(\dfrac{1}{100}-1\right)\). So sánh B với \(\dfrac{-11}{21}\)
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
1) Cho A= \(\dfrac{1}{\sqrt{100}}+\dfrac{1}{\sqrt{101}}+\dfrac{1}{\sqrt{102}}+\dfrac{1}{\sqrt{103}}+\dfrac{1}{\sqrt{104}}\)
và B= \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{100^2}\right)\)
So sánh A và B.
Thu gọn các biểu thức sau
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{299}\right)\)
C = \(-\dfrac{7}{4}.\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{333333}{424242}\right)\)
a, \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}.\sqrt{\dfrac{49}{4}}\right)\): \(\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]\): \(\dfrac{1704}{445}\)
b, \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)
c, \(\left(1-\dfrac{1}{2}\right)\)x\(\left(1-\dfrac{1}{3}\right)\)x.....x\(\left(1-\dfrac{1}{n+1}\right)\) (n ϵ N)
d, -66 x \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)\) + 124 x -37 + 63 x -124
e, \(\dfrac{7}{4}\) x \(\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
a, 1 + \(\dfrac{1}{2}\).(1+2)+\(\dfrac{1}{3}\).(1+2+3)+...+\(\dfrac{1}{16}\).(1+2+3+...+16)
b, \(\left[\left(\dfrac{2}{196}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]\):\(\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
c, \(\dfrac{\dfrac{1}{2}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\)x\(\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}\)+\(\dfrac{5}{8}\)
d, \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}\)+\(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
1) A = \(\left(-\dfrac{25}{27}-\dfrac{31}{42}\right)-\left(\dfrac{-7}{27}-\dfrac{3}{42}\right)\)
2) B = \(\dfrac{10\dfrac{3}{10}-\left(9,5-0,25\times18\right)\div0,5}{1\dfrac{1}{5}-1\dfrac{1}{2}}\)
3) C = \(\dfrac{3}{49}\times\dfrac{19}{2}-\dfrac{3}{49}\times\dfrac{5}{2}-\left(\dfrac{1}{20}-\dfrac{1}{4}\right)^2\times\left(\dfrac{-1}{2}-\dfrac{193}{14}\right)\)
a) \(\dfrac{\left(x+\dfrac{3}{4}\right)\cdot\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right)\cdot\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
b)\(\dfrac{\left(5-\dfrac{2}{7}\right)\cdot\dfrac{7}{9}\cdot\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
Tính giá trị biểu thức
A=\(\dfrac{\left(1+17\right)\cdot\left(1+\dfrac{17}{2}\right)\cdot\left(1+\dfrac{17}{3}\right)....\left(1+\dfrac{17}{19}\right)}{\left(1+19\right)\cdot\left(1+\dfrac{19}{2}\right)\cdot\left(1+\dfrac{19}{3}\right)....\left(1+\dfrac{19}{17}\right)}\)