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A = 2^15.(1+2^2+2^4) = 2^15.21 chia hết cho 21
=> ĐPCM
k mk nha
\(A=2^{15}+2^{17}+2^{19}\)
\(=2^{15}+2^{15}.2^2+2^{15}.2^4\)
\(=2^{15}\left(1+4+16\right)\)
\(=2^{15}.21⋮21\)
\(\Rightarrow A⋮21\)
Vậy \(A⋮21\)
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\(A=2^{15}+2^{17}+2^{19}=2^{14}\left(2+2^3+2^5\right)=2^{14}.42⋮21\)
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\(A=2+2^2+......+2^{59}+2^{60}\)
\(A=2\left(1+2\right)+....+2^{59}\left(1+2\right)\)
\(A=2\cdot3+...+2^{59}\cdot3⋮3\)
\(2+2^2+2^3+....+2^{58}+2^{59}+2^{60}\)
\(=2\left(1+2+4\right)+....+2^{58}\left(1+2+4\right)\)
\(=2\cdot7+.....+2^{58}\cdot7⋮7\)
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a. = 1/20 + 5 - 1/2
= 101/20 - 1/2
= 91/20
b. = ( 6/15 - 3/5) - ( 7/8 + 2/16) + 3
= -1/5 - 1 + 3
= 9/5
c. = 15/7 . ( 3/5 - 8/5)
= 15/7 . ( -1)
= - 15/7
e. = -14/9 - 3/9
= -17/9
f. = 19/21 . ( 15/17 + 2/17) + 13/21
= 19/21 . 1 + 13/21
= 32/21
g. = 43/12 : 2 + 5/24
= 43/24 + 5/24
= 2
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a, \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}-\dfrac{20}{15}+\dfrac{3}{7}\)
= \(\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{3}{7}\right)-\dfrac{20}{15}\)
= 1 + \(\dfrac{16}{21}-\dfrac{20}{15}\)
= \(\dfrac{3}{7}\)
b, \(\dfrac{27}{25}+\dfrac{4}{21}-\dfrac{2}{25}+\dfrac{17}{21}-\dfrac{1}{2}\)
= \(\left[\dfrac{27}{25}+\left(-\dfrac{2}{25}\right)\right]+\left(\dfrac{4}{21}-\dfrac{7}{21}\right)-\dfrac{1}{2}\)
= 1 + \(\left(\dfrac{-1}{7}\right)-\dfrac{1}{2}\) = \(\dfrac{5}{14}\)
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\(2^{41}+4^{21}+8^{15}=2^{41}+\left(2^2\right)^{21}+\left(2^3\right)^{15}=2^{41}+2^{42}+2^{45}=2^{41}\left(1+2+2^4\right)=2^{41}.19\) chia hết cho 19(đpcm)
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\(\frac{1}{5}A=\frac{1}{5}.\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{20}}\right)\)
\(\Rightarrow\frac{1}{5}A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{20}}\)
\(\Rightarrow\frac{1}{5}A-A=\left(\frac{1}{5^2}+...+\frac{1}{5^{21}}\right)-\left(\frac{1}{5}+...+\frac{1}{5^{20}}\right)\)
\(-\frac{4}{5}A=\frac{1}{5^{21}}-\frac{1}{5}\)
\(\Rightarrow A=\left(\frac{1}{5^{21}}-\frac{1}{5}\right):\left(-\frac{4}{5}\right)\)
các câu còn lại tương tự thôi
B1 c2
dùng xích ma \(\text{∑}^{20}_1\left(\frac{1}{5^x}\right)=0,25=\frac{1}{4}\)
chỗ phía dưới là 1 nha nó bị che