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a) A = {\(\dfrac{1}{n\left(n+1\right)}\)| \(n\in\mathbb{N},1\le n\le5\)}
b) B = {\(\dfrac{1}{n^2-1}\)|\(n\in\mathbb{N},2\le n\le6\)\(\)}
Ta có:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}>4\cdot\dfrac{1}{16}=\dfrac{1}{4}\)
\(\dfrac{1}{17}+\dfrac{1}{18}+\dfrac{1}{19}+\dfrac{1}{20}>4\cdot\dfrac{1}{20}=\dfrac{1}{5}\)
=>\(\dfrac{1}{13}+\dfrac{1}{14}+...+\dfrac{1}{20}>\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\)
=>A>\(\dfrac{1}{12}+\dfrac{9}{20}\)
\(\dfrac{1}{12}>\dfrac{1}{20}\)
=>\(A>\dfrac{1}{20}+\dfrac{9}{20}=\dfrac{1}{2}\)
Vậy...
\(A=\dfrac{1}{2}+\dfrac{3-2}{3.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{100.99}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}\)
\(A=\dfrac{99}{100}\)
\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{2007.2009}+\dfrac{2}{2009..2011}\)
\(2B=\dfrac{3-1}{1.3}+\dfrac{5-3}{3,5}+...+\dfrac{2009-2007}{2009.2007}+\dfrac{2011-2009}{2011.2009}\)
\(2B=\dfrac{3}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\)
\(2B=1-\dfrac{1}{2011}\)
\(2B=\dfrac{2010}{2011}\)
\(B=\dfrac{2010}{4022}\)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
a: \(=\left(\dfrac{1}{15}+\dfrac{14}{15}\right)+\left(\dfrac{9}{10}-2-\dfrac{11}{9}\right)+\dfrac{1}{157}\)
\(=1+\dfrac{1}{157}+\dfrac{81-180-110}{90}\)
\(=\dfrac{158}{157}+\dfrac{-209}{90}\simeq-1.315\)
b: \(=\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{2}{6}\)
=1/3-1/3
=0
c: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2015\cdot2017}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
=2016/2017
bài giải:
đặt biểu thức bằng A
=> A= \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
ta thấy:\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< 3.\dfrac{1}{13}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< 3.\dfrac{1}{61}\)
=> A<\(\dfrac{1}{5}+\dfrac{3}{13}+\dfrac{3}{61}\)<\(\dfrac{1}{2}\)
=> đpcm.
a: \(=\left(\dfrac{-48}{12}+\dfrac{-8}{12}+\dfrac{21}{12}\right)\cdot\dfrac{-12}{13}\)
\(=\dfrac{-35}{12}\cdot\dfrac{-12}{13}=\dfrac{35}{13}\)
b: \(=\dfrac{-3}{6}+\dfrac{5}{6}-\dfrac{312}{100}+\dfrac{51}{10}\)
\(=\dfrac{1}{3}-\dfrac{312}{100}+\dfrac{51}{10}=\dfrac{347}{150}\)
c: \(=\left(\dfrac{48}{300}+\dfrac{175}{300}-\dfrac{135}{100}\right)\cdot\dfrac{5}{2}+\dfrac{1}{4}\)
\(=\dfrac{88}{300}\cdot\dfrac{5}{2}+\dfrac{1}{4}=\dfrac{59}{60}\)
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
Bài 2:
a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)
d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
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