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Ta có \(\frac{3n+2}{n-1}=\frac{3\left(n-1\right)+5}{n-1}=3+\frac{5}{n-1}\)
để A có giá trị nguyên thì 5 phải chia hết cho n-1 hay n-1 là ước của 5
Ư(5)={5,1,-1,-5}
\(\Rightarrow\)n={6,2,0,-4}
gọi số cần tìm là A,Ta có: A+2CHIA HẾT CHO 3,4,5,6 HAY A+2 là bội chung của 3,4,5,6
BCNN(3,4,5,6)=60
\(\Rightarrow A+2=60.n\Rightarrow n=1,2,3,4,.... \)
lần lượt thử các số n.
Ta thấy n=7 thì A=418 chia hết cho 11
vậy số nhỏ nhất là 418
1)Ta có:\(2^{60}=\left(2^3\right)^{20}=8^{20}\)
\(3^{40}=\left(3^2\right)^{20}=9^{20}\)
Vì \(8^{20}< 9^{20}\Rightarrow2^{60}< 3^{40}\)
2)Gọi d là ƯCLN(n+3,2n+5)(d\(\in N\)*)
Ta có:\(n+3⋮d,2n+5⋮d\)
\(\Rightarrow2n+6⋮d,2n+5⋮d\)
\(\Rightarrow\left(2n+6\right)-\left(2n+5\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vì ƯCLN(n+3,2n+5)=1\(\RightarrowƯC\left(n+3,2n+5\right)=\left\{1,-1\right\}\)
3)\(A=5+5^2+5^3+5^4+...+5^{98}+5^{99}\)(có 99 số hạng)
\(A=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{97}+5^{98}+5^{99}\right)\)(có 33 nhóm)
\(A=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{97}\left(1+5+5^2\right)\)
\(A=5\cdot31+5^4\cdot31+...+5^{97}\cdot31\)
\(A=31\left(5+5^4+...+5^{97}\right)⋮31\left(đpcm\right)\)
6)Đặt \(A=2^1+2^2+2^3+...+2^{100}\)
\(2A=2^2+2^3+2^4+...+2^{101}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2^1+2^2+2^3+...+2^{100}\right)\)
\(A=2^{101}-2\)
\(\Rightarrow2^1+2^2+2^3+...+2^{100}-2^{101}=2^{101}-2-2^{101}=-2\)
Bài 1:
a) \(\frac{\left(-3\right)}{16}+\frac{1}{15}=\frac{-45}{240}+\frac{16}{240}\)
\(=\frac{-29}{240}\)
b)\(\frac{\left(-15\right)}{24}-\frac{\left(-2\right)}{6}=\frac{\left(-15\right)}{24}-\frac{-8}{24}\)
\(=\frac{-7}{24}\)
c) \(\frac{\left(-16\right)}{18}\cdot\frac{36}{\left(-40\right)}=\frac{\left(-8\right)}{9}\cdot\frac{\left(-9\right)}{10}\)
\(=\frac{\left(-80\right)}{90}\cdot\frac{\left(-81\right)}{90}\)
\(=\frac{4}{5}\)
d)\(\frac{\left(-17\right)}{30}:\frac{34}{60}=\frac{\left(-17\right)}{30}:\frac{17}{30}\)
\(=\frac{\left(-17\right)}{30}\cdot\frac{30}{17}\)
\(=-1\)
Bài 2:
a) \(1\frac{3}{5}+2\frac{1}{6}=\frac{8}{5}+\frac{13}{6}=\frac{48}{30}+\frac{65}{30}\)
\(=\frac{113}{30}\)
b) \(3\frac{1}{7}-1\frac{1}{8}=\frac{22}{7}-\frac{9}{8}=\frac{176}{56}-\frac{63}{56}\)
\(=\frac{113}{56}\)
c) \(3\frac{1}{6}\cdot2\frac{1}{4}=\frac{19}{6}\cdot\frac{9}{4}=\frac{57}{8}\)
d) \(4\frac{1}{5}:3\frac{6}{7}=\frac{21}{5}:\frac{27}{7}=\frac{21}{5}\cdot\frac{7}{27}\)
\(=\frac{49}{45}\)
a \(\frac{-3}{16}+\frac{1}{15}=\frac{-45}{240}+\frac{16}{240}=\frac{-29}{240}\)
b \(\frac{-15}{24}-\frac{-2}{6}=\frac{-15}{24}-\frac{-8}{24}=\frac{-7}{24}\)
c \(\frac{-16}{18}.\frac{36}{-40}=\frac{4}{5}\)
d \(\frac{-17}{30}:\frac{34}{60}=\frac{-17}{30}.\frac{60}{34}=-1\)
bai 2
\(1\frac{3}{5}+2\frac{1}{6}=\frac{8}{5}+\frac{13}{6}=\frac{113}{30}\)
\(3\frac{1}{7}-1\frac{1}{8}=\frac{22}{7}-\frac{9}{8}=\frac{113}{56}\)
c \(3\frac{1}{6}.2\frac{1}{4}=\frac{19}{6}.\frac{9}{4}=\frac{57}{8}\)
d \(4\frac{1}{5}:3\frac{6}{7}=\frac{21}{5}:\frac{27}{7}=\frac{21}{5}.\frac{7}{27}=\frac{147}{135}\)
1) K = D. 10 000 + Q
=> K-Q = D.10 000
=> 2015(K-Q) + 2016D = 2015.D.10 000 + 2016D =20152016.D
Vậy 2015(K-Q) + 2016D chia cho D = 20152016D:D = 20152016
2) \(A=\frac{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}\right)}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=\)
\(A=\frac{\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(1+\frac{1}{2}+\frac{1}{3}\right)}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=\)
\(=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}=1\)