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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
Bài 3:
Để A là số nguyên thì \(n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
a. 32 = 25 => n thuộc tập 1; 2; 3; 4
b. \(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{4}+\frac{2}{3}=\frac{11}{12}\)
\(\Rightarrow x=\frac{12}{11}\)
c. p nguyên tố => \(p\ge2\) => 52p luôn có dạng A25
=> 52p+2015 chẵn
=> 20142p + q3 chẵn
Mà 20142p chẵn => q3 chẵn => q chẵn => q = 2
=> 52p + 2015 = 20142p+8
=> 52p+2007 = 20142p
2014 có mũ dạng 2p => 20142p có dạng B6
=> 52p = B6 - 2007 = ...9 (vl)
(hihi câu này hơi sợ sai)
d. \(17A=\frac{17^{19}+17}{17^{19}+1}=1+\frac{16}{17^{19}+1}\), \(17B=\frac{17^{18}+17}{17^{18}+1}=1+\frac{16}{17^{18}+1}\)
\(17^{19}+1>17^{18}+1\Rightarrow\frac{16}{17^{19}+1}< \frac{16}{17^{18}+1}\)
\(\Rightarrow17A< 17B\)
\(\Rightarrow A< B\)
2) Để A là nguyên thì n - 1 là ước nguyên của 2
\(n-1=1\Rightarrow n=2\)
\(n-1=2\Rightarrow n=3\)
3) Ta gọi M là \(\dfrac{12}{5^{2012}}\)
\(M=\dfrac{5.12}{5^{2012}.5}=\dfrac{60}{5^{2013}}\)
\(\Rightarrow\) \(A=\dfrac{60}{5^{2013}}+\dfrac{18}{5^{2013}}=\dfrac{78}{5^{2013}}\)
Ta gọi Q là \(\dfrac{18}{5^{2012}}\)
\(Q=\dfrac{18}{5^{2012}}=\dfrac{18.5}{5^{2012}.5}=\dfrac{90}{5^{2013}}\)
\(\Rightarrow\) \(B=\dfrac{90}{5^{2013}}+\dfrac{12}{5^{2013}}=\dfrac{102}{5^{2013}}\)
\(\dfrac{90}{5^{2013}}< \dfrac{102}{5^{2013}}\Rightarrow A< B\)
Ai thấy đúng thì ủng hộ mink, thấy sai góp ý nha !!!
\(1)\dfrac{1}{5}+\dfrac{2}{30}+\dfrac{121}{156}\le x\le\dfrac{1}{2}+\dfrac{156}{72}+\dfrac{1}{3}\)
\(\dfrac{156}{780}+\dfrac{26}{780}+\dfrac{605}{780}\le x\le\dfrac{3}{6}+\dfrac{13}{6}+\dfrac{2}{6}\)
\(\dfrac{787}{780}\le x\le2\)
\(\Rightarrow x\in\left\{2\right\}\)
Câu 2:
\(N=\dfrac{2a+9+5a+17-3a-4a-23}{a+3}=\dfrac{3}{a+3}\)
Để N là số tự nhiên thì \(\left\{{}\begin{matrix}a>-3\\a+3\in\left\{1;-1;3;-3\right\}\end{matrix}\right.\Leftrightarrow a\in\left\{-2;0\right\}\)
Câu 1:
a, \(\left|-5\right|=5\)
b, \(\left|10\right|=10\)
c, \(\left|-5\right|-\left|10\right|=5-10=-5\)
d, -15.30= -450
Câu 2:
a, Ta có: \(\dfrac{10}{21}.\dfrac{14}{25}=\dfrac{10.14}{21.25}=\dfrac{5.2.7.2}{3.7.5.5}=\dfrac{2.2}{3.5}=\dfrac{4}{15}\)
c, Ta có: \(-\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{-5.2+3.3}{12}=\dfrac{-10+9}{12}=\dfrac{-1}{12}\)
d, \(\dfrac{11}{17}.\dfrac{3}{2017}+\dfrac{11}{17}.\dfrac{2014}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}\left(\dfrac{3}{2017}+\dfrac{2014}{2017}\right)-1\dfrac{11}{17}\)
\(=\dfrac{11}{17}.\dfrac{2017}{2017}-1\dfrac{11}{17}=\dfrac{11}{17}-1-\dfrac{11}{17}=-1\)
Câu 7: a, Để A có nghĩa khi \(x+2\ne0\) \(\Leftrightarrow x=-2\)
b, Ta có: \(A=2\)
<=> \(\dfrac{x-1}{x+2}=2\)
<=> \(\dfrac{x-1}{x+2}-2=0\)
<=> \(\dfrac{x-1}{x+2}-\dfrac{2x+4}{x+2}=0\)
<=> \(\dfrac{x-1-2x-4}{x+2}=0\)
<=> \(\dfrac{-x-5}{x+2}=0\)
<=> -x-5=0
<=> -x=5
<=> x= -5
1) Tìm tập hợp A sao cho các số nguyên a sao cho:
=> \(\dfrac{1}{2}\) +\(\dfrac{1}{34}\) \(\le\) \(\dfrac{a}{17}\) <\(\dfrac{15}{17}\) - \(\dfrac{3}{17}\)
\(\dfrac{17}{34}\)+\(\dfrac{1}{34}\)\(\le\)\(\dfrac{a}{34}\)<\(\dfrac{12}{17}\)
\(\dfrac{18}{34}\) \(\le\)\(\dfrac{a}{34}\)<\(\dfrac{24}{34}\)
=> a \(\in\) {18; 19; 20; 21; 22; 23 }
2)
Để A là số nguyên thì 2 phải chia hết cho n-1
=> n-1 \(\in\) ước của 2
=> n-1\(\in\) {1;-1;2;-2}
=> n\(\in\) {-1; 0; 2; 3}