2 chia 2/3
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a) \(\left(1+2+2^2+...+2^7\right)\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)
\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)
\(=3+2^2.3+...+2^6.3\)
\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)
a) Đặt A = 1 + 2 + 22 + 23 + ... + 27
Ta có:
A = 1 + 2 + 22 + 23 + ... + 27
\(\Rightarrow\)2A = 2 + 22 + 23 + 24 + ... + 28
\(\Rightarrow\)A = 28 - 1 = 255
Vì 255\(⋮\)3\(\Rightarrow\)2 + 22 + 23 + 24 + ... + 28\(⋮\)3
\(\Rightarrow\)ĐPCM
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{3}{5}\)
`=`\(\dfrac{9}{12}+\dfrac{8}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{17}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{85}{60}+\dfrac{36}{60}\)
`=`\(\dfrac{121}{60}\)
`b)`
\(\dfrac{1}{2}\cdot\dfrac{9}{13}\div\dfrac{27}{26}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{9}{13}\cdot\dfrac{26}{27}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{2}{3}\)
`=`\(\dfrac{1}{3}\)
`c)`
\(\dfrac{2}{7}\cdot\dfrac{1}{9}+\dfrac{2}{7}\cdot\dfrac{2}{9}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\left(\dfrac{1}{9}+\dfrac{2}{9}\right)+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{1}{3}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
`=`\(\dfrac{1}{3}\cdot1=\dfrac{1}{3}\)
`d)`
\(11\div\dfrac{5}{2}+11\div\dfrac{7}{3}+11\div\dfrac{35}{6}\)
`=`\(11\cdot\dfrac{2}{5}+11\cdot\dfrac{3}{7}+11\cdot\dfrac{6}{35}\)
`=`\(11\cdot\left(\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{6}{35}\right)\)
`=`\(11\cdot1=11\)
a) 3/4 + 2/3 + 3/5 = 45/60 + 40/60 + 36/60 = 121/60
b) 1/2 x 9/13 : 27/26 = 9/26 x 26/27 = 1/3
c) 2/7 x 1/9 + 2/7 x 2/9 + 1/3 x 5/7 = 2/7 x (1/9 + 2/9) + 5/21 = 2/7 x 1/3 + 5/21 = 2/21 + 5/21 = 1/3
d) 11 : 5/2 + 11 : 7:3 + 11 : 35/6 = 11 x (2/5 + 3/7 + 6/35) = 11 x 1 = 11
1) Ta có: \(2⋮n-3\)
\(\Leftrightarrow n-3\inƯ\left(2\right)\)
\(\Leftrightarrow n-3\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{4;2;5;1\right\}\)
Vậy: \(n\in\left\{4;2;5;1\right\}\)
2) Ta có: \(n+2⋮n-3\)
\(\Leftrightarrow n-3+5⋮n-3\)
mà \(n-3⋮n-3\)
nên \(5⋮n-3\)
\(\Leftrightarrow n-3\inƯ\left(5\right)\)
\(\Leftrightarrow n-3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{4;2;8;-2\right\}\)
Vậy: \(n\in\left\{4;2;8;-2\right\}\)
2 : \(\dfrac{2}{3}\)
= 2 x \(\dfrac{3}{2}\)
= 3
2:2/3 =3