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b) Ta có:
\(B=\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{1}{2016}\)
\(\Rightarrow B=\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{1}{2016}+1\right)+1\)
\(\Rightarrow B=\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2016}+\frac{2017}{2017}\)
\(\Rightarrow B=2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}}{2017\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}\right)}=\frac{1}{2017}\)
Vậy \(\frac{A}{B}=\frac{1}{2017}\)
Bài 1:
\(A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{99}}\)
\(\Leftrightarrow\dfrac{1}{5}A=\dfrac{1}{5^2}+\dfrac{1}{5^3}+\dfrac{1}{5^4}+...+\dfrac{1}{5^{100}}\)
Lây vế trừ vế, ta được:
\(A-\dfrac{1}{5}A=\dfrac{4}{5}A\)
\(\dfrac{4}{5}A=\dfrac{1}{5}-\dfrac{1}{5^{100}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{5}-\dfrac{1}{5^{100}}}{\dfrac{4}{5}}=\dfrac{\dfrac{1}{5}.\left(1-\dfrac{1}{5^{99}}\right)}{\dfrac{1}{5}.4}=\dfrac{1-\dfrac{1}{5^{99}}}{4}\)
Vậy \(A=\dfrac{1-\dfrac{1}{5^{99}}}{4}\).
Chúc bạn học tốt!
Bài 2:
Có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\)
\(\Leftrightarrow B=273+...+3^{1986}.273\)
\(\Leftrightarrow B=273\left(1+...+1986\right)\)
Vì \(273⋮13\)
Nên \(B=273\left(1+...+1986\right)⋮13\)
Vậy \(B⋮13\)
Lại có:
\(B=3+3^3+3^5+...+3^{1991}\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Leftrightarrow B=\left(3+3^3+3^5+3^7\right)+...+3^{1984}\left(3+3^3+3^5+3^7\right)\)
\(\Leftrightarrow B=2460+...+3^{1984}.2460\)
\(\Leftrightarrow B=2460\left(1+...+3^{1984}\right)\)
Vì \(2460⋮41\)
Nên \(B=2460\left(1+...+3^{1984}\right)⋮41\)
Vậy \(B⋮41\).
Chúc bạn học tốt!
Gọi d là ước chung nguyên tố của 35n + 5 và 3n + 1
\(\Rightarrow\left\{\begin{matrix}35n+5⋮d\\3n+1⋮d\end{matrix}\right.\)
+) Vì : \(35n+5⋮d;3\in Z\Rightarrow3\left(35n+5\right)⋮d\) \(\Rightarrow105n+15⋮d\)
+) Vì : \(3n+1⋮d;35\in Z\Rightarrow35\left(3n+1\right)⋮d\Rightarrow105n+35⋮d\)
Mà : \(105n+15⋮d\)
\(\Rightarrow\left(105n+35\right)-\left(105n+15\right)⋮d\)
\(\Rightarrow105n+35-105n-15⋮d\Rightarrow20⋮d\)
\(\Rightarrow d\) là ước của 20
Mà : d là số nguyên tố \(\Rightarrow d\in\left\{2;5\right\}\)
Với d = 2 ; Mà : \(n\in Z\Rightarrow3n+1⋮̸\) 2 => loại
Với d = 5 : \(3n+1⋮5\Rightarrow4n-n+4-3⋮5\)
\(\Rightarrow4\left(n+1\right)-n-3⋮5\Rightarrow4\left(n+1\right)+\left(n-3\right)⋮5\)
\(\Rightarrow n-3⋮5\Rightarrow n-3=5k\Rightarrow n=5k+3\left(k\in Z\right)\)
Thử lại , ta có :
\(35n+5=35\left(5k+3\right)+5=175k+105=5\left(35k+21\right)⋮5\)
\(3n+1=3\left(5k+3\right)+1=15k+9+1=15k+10=5\left(3k+2\right)⋮5\)
Vậy n = 5k + 3 thì phân số trên rút gọn được
p/s : các câu khác làm tương tự
\(H=\left(9\frac{3}{8}+7\frac{3}{8}\right)+4,03=16\frac{3}{8}+4,03=16,375+4,03=20,405\)
\(I=10101.\left(\frac{5}{111111}+\frac{2,5}{111111}-\frac{4}{111111}\right)=10101.\frac{3,5}{111111}=\frac{7}{22}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+....+\dfrac{3}{59.61}\)
\(S=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{59}-\dfrac{1}{61}\)
\(S=\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{1}{5}-\dfrac{1}{61}\)
\(S=\dfrac{56}{305}\)
Vậy S = \(\dfrac{56}{305}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
\(S=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}.\dfrac{56}{305}=\dfrac{84}{305}\)
-1/6+1/6+0
-1/3+1/3+0
-1/2+1/2+0
mk bổ sung thêm nha:
\(\frac{-1}{6}+\frac{-1}{3}+\frac{1}{2}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{-1}{2}\)
Cộng thêm 3 cách của bạn Linh là đủ nhé :)