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Câu 2:
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\)
\(=2014\left(1+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\)
\(=2014\left(1+\dfrac{1}{2\left(2+1\right)}.2+\dfrac{1}{3\left(3+1\right)}.2+...+\dfrac{1}{2013\left(2013+1\right)}.2\right)\)
\(=2014\left(\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2013.2014}\right)\)
\(=4028\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2013.2014}\right)\)
Bạn tự tính nốt nhé
1)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\left(1\right)\)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2011\cdot2012}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\\ =\dfrac{1}{1}-\dfrac{1}{2012}< 1\left(2\right)\)
Từ (1) và (2) ta có: A < 1
2)
\(A=2014+\dfrac{2014}{1+2}+\dfrac{2014}{1+2+3}+...+\dfrac{2014}{1+2+3+...+2013}\\ =2014\cdot\left(\dfrac{1}{1}+\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+2013}\right)\\ =2014\cdot\left(\dfrac{1}{\left(1\cdot2\right):2}+\dfrac{1}{\left(2\cdot3\right):2}+\dfrac{1}{\left(3\cdot4\right):2}+...+\dfrac{1}{\left(2013\cdot2014\right):2}\right)\\ =2014\cdot\left(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{2013\cdot2014}\right)\\ =2014\cdot2\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2013\cdot2014}\right)\\ =4028\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\right)\\ =4028\cdot\left(1-\dfrac{1}{2014}\right)\\ =4028\cdot\dfrac{2013}{2014}\\ =4026\)
3)
Để A là số nguyên thì \(6n+42⋮6n\Rightarrow42⋮6n\Rightarrow6n\inƯ\left(42\right)\)
\(Ư\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)
| 6n | 1 | 2 | 3 | 6 | 7 | 14 | 21 | 42 |
| n | \(\dfrac{1}{6}\) | \(\dfrac{1}{3}\) | \(\dfrac{1}{2}\) | 1 | \(\dfrac{7}{6}\) | \(\dfrac{7}{3}\) | \(\dfrac{7}{2}\) | 7 |
Vì n là số tự nhiên nên n = 1 hoặc n = 7
4)
\(A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\cdot\left(17^{17}+1\right)}{17\cdot\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)
Vậy A<B
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
Bài này có rất nhiều cách lm nhé!
Ta có : A = \(\dfrac{17^{18}+1}{17^{19}+1}\) => 17A = \(\dfrac{17^{19}+17}{17^{19}+1}\) = \(1+\dfrac{16}{17^{19}+1}\)
B = \(\dfrac{17^{17}+1}{17^{18}+1}\) => 17B = \(\dfrac{17^{18}+17}{17^{18}+1}\) = \(1+\dfrac{16}{17^{18}+1}\)
Vì \(\dfrac{16}{17^{19}+1}\) < \(\dfrac{16}{17^{18}+1}\) ( vì 1719 +1 > 1716+1 )
=> \(1+\dfrac{16}{17^{19}+1}\) < \(1+\dfrac{16}{17^{18}+1}\)
=> 17A < 17B
=> A < B ( vì 17 > 0)
Ta có :
\(A=\dfrac{17^{18}+1}{17^{19}+1}\)
17A= \(17\times\dfrac{17^{18}+1}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+17}{17^{19}+1}\)
\(17A=\dfrac{\left(17^{19}+1\right)+16}{17^{19}+1}\)
\(17A=\dfrac{17^{19}+1}{17^{19}+1}+\dfrac{16}{17^{19}+1}\)
\(17A=1+\dfrac{16}{17^{19}+1}\)
Lại có :
\(B=\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=17\times\dfrac{17^{17}+1}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}\)
\(17B=\dfrac{\left(17^{18}+1\right)+16}{17^{18}+1}\)
\(17B=\dfrac{17^{18}+1}{17^{18}+1}+\dfrac{16}{17^{18}+1}\)
\(17B=1+\dfrac{16}{17^{18}+1}\)
Mà : \(\dfrac{16}{17^{19}+1}< \dfrac{16}{17^{18}+1}\)
\(\Rightarrow1+\dfrac{16}{17^{19}+1}< 1+\dfrac{16}{17^{18}+1}\)
⇒ A < B
Vậy A < B
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)
\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)
\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)
\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)
Bài 1:
a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)
b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)
\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)
c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)
\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)
Ta có F < 1 nên
\(\dfrac{13580}{34569}< 1< \dfrac{13580+\left(-1\right)}{34569+\left(-1\right)}=\dfrac{13579}{34568}\)<\(\dfrac{13579}{34567}\)
Từ đó suy ra \(\dfrac{13580}{34569}< \dfrac{13579}{34567}\)hay\(\dfrac{13579}{34567}>\dfrac{13580}{34569}\)
Vậy E > F
a, Ta có:
\(-5< 0;\dfrac{1}{63}>0\Rightarrow-5< \dfrac{1}{63}\)
b, Ta có:
\(-\dfrac{18}{17}< -1;\dfrac{999}{-1000}>-1\Rightarrow-\dfrac{18}{17}< \dfrac{999}{-1000}\)
c, Ta có:
\(-\dfrac{17}{35}>-\dfrac{1}{2};-\dfrac{43}{85}< -\dfrac{1}{2}\Rightarrow-\dfrac{17}{35}>-\dfrac{43}{85}\)
d, Ta có:
\(-0,76=-\dfrac{19}{25}\)
Vì \(25< 28\Rightarrow\dfrac{19}{25}>\dfrac{19}{28}\Rightarrow-\dfrac{19}{25}< -\dfrac{19}{28}\)
Chúc bạn học tốt!!!
a: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà 17^19+1>17^18+1
nên A<B
b: \(2C=\dfrac{2^{2021}-2}{2^{2021}-1}=1-\dfrac{1}{2^{2021}-1}\)
\(2D=\dfrac{2^{2022}-2}{2^{2022}-1}=1-\dfrac{1}{2^{2022}-1}\)
2^2021-1<2^2022-1
=>1/2^2021-1>1/2^2022-1
=>-1/2^2021-1<-1/2^2022-1
=>C<D
cho mình bài c với đc ko?mình ko bik làm😫😖