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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!
Ta có : \(\overline{abcdeg}=\overline{ab}.1000+\overline{cd}.100+\overline{eg}\)
\(=9999.\overline{ab}+\overline{ab}+99.\overline{cd}+\overline{cd}+\overline{eg}\)
\(=\left(9999.\overline{ab}+99.\overline{cd}\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
Vì : \(9999.\overline{ab}+99.\overline{cd}⋮11\) và \(\overline{ab}+\overline{cd}+\overline{eg}⋮11\)
\(\Rightarrow\overline{abcdeg}⋮11\left(đpcm\right)\)
Ta có:
\(\overline{abcdeg}=\overline{ab}.10000+\overline{cd}.100+\overline{eg}\)
\(=\overline{ab}.9999+\overline{ab}+\overline{cd}.99+\overline{cd}+\overline{eg}\)
\(=\overline{ab}.11.909+\overline{cd}.11.9+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
\(=11\left(\overline{ab}.909+\overline{cd}.9\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
Vì \(11\left(\overline{ab}.909+\overline{cd}.9\right)⋮11\) và \(\overline{ab}+\overline{cd}+\overline{eg}⋮11\)
nên \(\overline{abcdeg}⋮11\)
Vậy nếu \(\overline{ab}+\overline{cd}+\overline{eg}⋮11\) thì \(\overline{abcdeg}⋮11\) (đpcm)
Chứng Minh:C=\(3^0+3^2+3^4+...+3^{2002}⋮7\)
Nhân C với \(3^2\)ta có:
\(9S=3^2+3^4+3^6+...+3^{2004}\)
\(\Rightarrow9S-S=\left(3^2+3^4+...+3^{2004}\right)-\left(3^0+3^2+3^4+...+3^{2002}\right)\)
\(\Rightarrow8S=3^{2004}-1\)
\(\Rightarrow S=\dfrac{3^{2004}-1}{8}\)
Chứng minh:
Ta có:\(3^{2004}-1=\left(3^6\right)^{334-1}=\left(3^6-1\right).a=7.104.a\)
\(\)UCLN(7;8)=1
\(\Rightarrow S⋮7\)
Sửa lại 1 chút!
Chứng minh: C= \(3^0+3^2+3^4+3^6+...+3^{2002}\) chia hết cho 7
\(4\dfrac{1}{3}.\dfrac{4}{9}+13\dfrac{2}{3}.\dfrac{4}{9}\)\(=\dfrac{4}{9}\left(4\dfrac{1}{3}+13\dfrac{2}{3}\right)=\dfrac{4}{9}.18=8\)
\(5\dfrac{1}{4}.\dfrac{3}{8}+10\dfrac{3}{4}.\dfrac{3}{8}=\dfrac{3}{8}\left(5\dfrac{1}{4}+10\dfrac{3}{4}\right)=\dfrac{3}{8}.16=6\)
(x+1/4-1/3).(13/6-1/4)=7/46
(x+1/4-1/3).23/12=7/46
(x+1/4-1/3)=7/46:23/12
(x+1/4-1/3)=7/46.12/23
(x+1/4-1/3)=42/529
x+1/4=42/529+1/3
x+1/4=655/1587
x=655/1587-1/4
x=1033=/6348
vậy x=1033/6348
a) \(27^{11}=\left(3^3\right)^{11}=3^{33}\)
\(81^8=\left(3^4\right)^8=3^{24}\)
\(\Rightarrow27^{11}>81^8\)
b) \(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
\(\Rightarrow5^{36}>11^{24}\)
c) \(5^{23}=5\cdot5^{22}\)
Ta có: \(6>5;5^{22}=5^{22}\)
\(\Rightarrow5^{23}< 6\cdot5^{22}\)