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Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
Bài 1:
a)\(\dfrac{2\cdot\left(-3\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}=\dfrac{2\cdot3\cdot3\cdot2\cdot5}{2\cdot2\cdot13\cdot2}=\dfrac{45}{26}\)
b)\(\dfrac{15\cdot8+15\cdot4}{12\cdot3}=\dfrac{15\cdot12}{12\cdot13}=\dfrac{15}{13}\)
Bài 1 :
a, \(\dfrac{2.\left(-13\right).9.10}{\left(-3\right).4.\left(-5\right).26}=\dfrac{-2.13.3.3.2.5}{3.2.2.5.2.13}=\dfrac{-3}{2}\) b, \(\dfrac{15.8+15.4}{12.3}=\dfrac{15\left(8+4\right)}{12.3}=\dfrac{3.5.12}{12.3}=5\)
Bài 1:
1,\(\dfrac{-14}{-52}\) =\(\dfrac{-14:\left(-2\right)}{-52:\left(-2\right)}\)=\(\dfrac{7}{26}\)
2,\(\dfrac{-36}{-10}\)=\(\dfrac{36:\left(-2\right)}{-10:\left(-2\right)}\)=\(\dfrac{-18}{5}\)
3,\(\dfrac{-35}{-156}\)=\(\dfrac{-35:\left(-1\right)}{-156:\left(-1\right)}\)=\(\dfrac{35}{156}\)
4,\(\dfrac{-360}{480}\)=\(\dfrac{-360:120}{480:120}\)=\(\dfrac{-3}{4}\)
5,\(\dfrac{3.5}{20.33}\)=\(\dfrac{3.5}{4.5.3.11}\)=\(\dfrac{1}{44}\)
6,\(\dfrac{-4.56}{16.32}=\dfrac{-4.7.8}{4.4.8.4}=\dfrac{-7}{16}\)
7,\(\dfrac{2.3.16}{24.21.8}=\dfrac{2.3.4.4}{6.4.7.3.2.4}=\dfrac{1}{42}\)
8,\(\dfrac{8.6-8.10}{16}=\dfrac{8.\left(6-10\right)}{8.2}=-2\)
9,\(\dfrac{18.9-18}{2-20}=\dfrac{18.\left(9-1\right)}{-18}=-8\)
a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
A = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}\)
B = \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{49.51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
B = \(\dfrac{1}{2}-\dfrac{1}{51}=\dfrac{51}{102}-\dfrac{2}{102}=\dfrac{49}{102}\)
\(\frac{5^3.2.3^2.5.2^6}{5^{10}.3^2.2^{13}}=\frac{5^4.2^7.3^2}{5^{10}.3^2.2^{13}}=\frac{1}{5^6.2^6}=\frac{1}{10^6}\)
\(\frac{18\left(27-23\right)}{4\left(34-52\right)}=\frac{18.4}{4.\left(-18\right)}=-1\)
a; \(\dfrac{3.5}{8.24}\) = \(\dfrac{3.5}{8.3.8}\) = \(\dfrac{5.3:3}{8.8.3:3}\) = \(\dfrac{5}{64}\)
b; \(\dfrac{8.6}{9.32}\)= \(\dfrac{8.2.3}{3.3.8.2.2}\) = \(\dfrac{8.2.3:\left(8.2.3\right)}{3.2.8.2.3:\left(8.2.3\right)}\) = \(\dfrac{1}{3.2}\) = \(\dfrac{1}{6}\)
b; \(\dfrac{6.5.12}{20.15}\) = \(\dfrac{6.5.4.3}{4.5.3.5}\) = \(\dfrac{6.3.4.5:\left(3.4.5\right)}{5.3.4.5:\left(3.4.5\right)}\) = \(\dfrac{6}{5}\)