cho B=3.5.7...99.101.(1+\(\dfrac{1}{3}\)+\(\dfrac{1}{5}\)+...+\(\dfrac{1}{101}\))
Chứng minh B chia hết cho 17
Chứng minh rằng số tự nhiên A chia hết cho 101 với:
A=1.2.3...99.100,(1\(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\))
Ta có: A=1.2.3.....99.100.(\(1+\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{99}+\dfrac{1}{100}\))
\(=1.2.3...100\left[\left(1+\dfrac{1}{100}\right)+\left(\dfrac{1}{2}+\dfrac{1}{99}\right)+......+\left(\dfrac{1}{50}+\dfrac{1}{51}\right)\right]\)
=>A= 1.2...100.\(\left[\dfrac{101}{100}+\dfrac{101}{2.99}+......+\dfrac{101}{50.51}\right]\)
=1.2.....100.101\(\left[\dfrac{1}{100}+\dfrac{1}{2.99}+.....+\dfrac{1}{50.51}\right]⋮101\)
Vậy A chia hết cho 101
Chứng minh:
a) \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
b) \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{2}{3}\)
a)
Ta thấy:
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
\(\dfrac{1}{12}< \dfrac{1}{10}\)
\(\dfrac{1}{13}< \dfrac{1}{10}\)
...
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 5\cdot\dfrac{1}{5}+8\cdot\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
b)
Ta thấy:
\(\dfrac{1}{101}>\dfrac{1}{300}\)
\(\dfrac{1}{102}>\dfrac{1}{300}\)
\(\dfrac{1}{103}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>200\cdot\dfrac{1}{300}=\dfrac{2}{3}\)
Vậy \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)
a)Tìm số nguyên sao cho 4n-5 chia hết cho n-3
b)Chứng minh rằng:
S=\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{2}\)
a) Giải:
Ta có: \(4n-5=4\left(n-3\right)+7\)
Để \(\left(4n-5\right)⋮\left(n-3\right)\Leftrightarrow7⋮n-3\)
\(\Rightarrow n-3\inƯ\left(7\right)\)
Mà \(Ư\left(7\right)\in\left\{\pm1;\pm7\right\}\)
Nên ta có bảng sau:
\(n-3\) | \(n\) |
\(1\) | \(4\) |
\(-1\) | \(2\) |
\(-7\) | \(-4\) |
\(7\) | \(10\) |
Vậy \(n=\left\{2;4;-4;10\right\}\)
b) Ta có:
\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)
\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)
Nhận xét:
\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)
\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)
Tổng 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{5}\) + ....... + \(\dfrac{1}{17}\) + \(\dfrac{1}{18}\) bằng \(\dfrac{a}{b}\) với \(\dfrac{a}{b}\) là phân số tối giải. Chứng minh rằng b chia hết cho 2431?
Bài 1:
1) Tìm x, biết: a. 52x-3 - 2 . 52 = 52. 3 b. /2x - 1/ = 5
2) Chứng minh rằng nếu 3a + 2b chia hết cho 17 thì 10a + b chia hết cho 17
Bài 2:
a. Tìm STN nhỏ nhất chia cho 36 dư 12 và chia cho 120 dư 60.
b. Tìm STN a, b, c, d nhỏ nhất sao cho: \(\dfrac{a}{b}\) = \(\dfrac{15}{21}\) ; \(\dfrac{b}{c}\) = \(\dfrac{9}{12}\) ; \(\dfrac{c}{d}\) = \(\dfrac{9}{11}\)
1, Chứng minh
a) A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
b) B=\(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+....+\dfrac{36}{25.26.27}< 3\)
a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760
Cho C=\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) Chứng minh rằng C > 5/8
Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)
\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)
\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)
\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)
\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)
hay \(C>\dfrac{5}{8}\)(đpcm)
1.Cho \(a,b,c,d\) là các số nguyên thỏa mãn \(a^3+b^3=2\left(c^3-d^3\right)\) . Chứng minh rằng a+b+c+d chia hết cho 3
2.Cho ba số dương a,b,c thỏa mãn abc=1. Chứng minh rằng \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
\(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}>\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{2019\cdot2020}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=\dfrac{1}{5}-\dfrac{1}{2020}=\dfrac{404-1}{2020}=\dfrac{403}{2020}>\dfrac{40}{2020}=\dfrac{20}{101}\left(1\right)\) \(A=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2019^2}< \dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+...+\dfrac{1}{2018\cdot2019}=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{2018}-\dfrac{1}{2019}=\dfrac{1}{4}-\dfrac{1}{2019}=\dfrac{2019-4}{4\cdot2019}=\dfrac{2015}{4\cdot2019}< \dfrac{2019}{4\cdot2019}=\dfrac{1}{4}\left(2\right)\) Từ (1) và (2) \(\Rightarrow\dfrac{20}{101}< A< \dfrac{1}{4}\)