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\(b,C=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{12}+...+\dfrac{1}{30}-\dfrac{1}{33}\\ =\dfrac{1}{3}-\dfrac{1}{33}\\ =\dfrac{11}{33}-\dfrac{1}{33}=\dfrac{10}{33}\)
a.F=\(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{2008.2010}\)
F=\(\dfrac{2.2}{2.4}\)+\(\dfrac{2.2}{4.6}\)+\(\dfrac{2.2}{6.8}\)+...+\(\dfrac{2.2}{2008.2010}\)
F=2.(\(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+\(\dfrac{2}{6.8}\)+...+\(\dfrac{2}{2008.2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2010}\))
F=2.(\(\dfrac{1}{2}\)-\(\dfrac{1}{2010}\))
F=\(\dfrac{1004}{1005}\)
\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)
\(=\dfrac{4}{9}-\dfrac{1}{5}\)
\(=\dfrac{11}{45}\)
\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)
\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)
\(S=.C.A.S.I.O.\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
Có gì ko hiểu thì cứ hỏi mình nha :)
Ta có: \(A=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2.2\frac{2}{4}+2.2\frac{2}{4.6}+2.2\frac{2}{6.8}+...+2.2\frac{2}{2008.2010}\)
\(=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{1}{2}-2.\frac{1}{2010}\)
\(=1-\frac{1}{1005}\)
\(=\frac{1004}{1005}\)
a) \(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{26\cdot31}\)
\(=1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\)
\(=1-\dfrac{1}{31}\)
\(=\dfrac{30}{31}\)
b) \(\dfrac{4}{11\cdot16}+\dfrac{4}{16\cdot21}+...+\dfrac{4}{61\cdot66}\)
\(=\dfrac{4}{5}\cdot\left(\dfrac{5}{11\cdot16}+\dfrac{5}{16\cdot21}+...+\dfrac{5}{61\cdot66}\right)\)
\(=\dfrac{4}{5}\cdot\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)
\(=\dfrac{4}{5}\cdot\left(\dfrac{1}{11}-\dfrac{1}{66}\right)\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{66}\)
\(=\dfrac{4}{66}\)
\(=\dfrac{2}{33}\)
a) A = 5²/(1.6) + 5²/(6.11) + ... + 5²/(26.31)
= 5.[5/(1.6) + 5/(6.11) + ...+ 5/(26.31)]
= 5.(1 - 1/6 + 1/6 - 1/11 + ... + 1/26 - 1/31)
= 5.(1 - 1/31)
= 5.30/31
= 150/31
b) B = 4/(11.16) + 4/(16.21) + ... + 4/(61.66)
= 4/5 .[5/(11.16) + 5/(16.21) + ... + 5/(61.66)]
= 4/5.(1/11 - 1/16 + 1/16 - 1/21 + ... + 1/61 - 1/66)
= 4/5.(1/11 - 1/66)
= 4/5 . 5/66
= 2/33
a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)
\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)
\(\Leftrightarrow x-4=25\)
\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)
b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)
\(\Leftrightarrow x\left(x+1\right)=18.4\)
\(\Leftrightarrow x\left(x+1\right)=72\)
vì \(72=8.9=\left(-8\right).\left(-9\right)\)
\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)
c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)
\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)
\(\Leftrightarrow2x+3-2x-8⋮x+4\)
\(\Leftrightarrow-5⋮x+4\)
\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)
_____
\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)
`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)
`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)
______
\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)
`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)
`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)
`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)
`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)
\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)
\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)
\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)
Vì \(\dfrac{1}{11}>\dfrac{1}{18}>\dfrac{1}{21}>\dfrac{1}{24}>\dfrac{1}{27}>\dfrac{1}{29}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}>\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}\)\(\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}=\dfrac{1}{11}.7=\dfrac{7}{11}\)
Ta có:
\(\dfrac{7}{11}=\dfrac{7.5}{11.5}=\dfrac{35}{55};\dfrac{4}{5}=\dfrac{4.11}{5.11}=\dfrac{44}{55}\)
\(Vì\) \(\dfrac{44}{55}>\dfrac{35}{55}\)
\(\Rightarrow\dfrac{4}{5}>\dfrac{7}{11}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< \dfrac{4}{5}\left(đpcm\right)\)
Ta thấy :
\(\dfrac{1}{4}+\dfrac{1}{11}< \dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{18}+\dfrac{1}{21}< \dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)
\(\dfrac{1}{24}+\dfrac{1}{27}< \dfrac{1}{24}+\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{1}{3}-\dfrac{1}{4}\)
\(\dfrac{1}{29}< \dfrac{1}{20}=\dfrac{1}{4}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{4}+\dfrac{1}{11}+\dfrac{1}{18}+\dfrac{1}{21}+\dfrac{1}{24}+\dfrac{1}{27}+\dfrac{1}{29}< 1-\dfrac{1}{5}=\dfrac{4}{5}\)
\(\Rightarrow dpcm\)
\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)
\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)
\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)
\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)
\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)
\(A=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\)
\(=2.\dfrac{502}{1005}=\dfrac{1004}{1005}\)