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\(\frac{1}{40}\)x\(\frac{1}{30}\)x\(\frac{1}{20}\)x\(\frac{1}{12}\)x\(\frac{1}{6}\)x\(\frac{1}{2}\)
= \(\frac{1}{40.30.20.12.6.2}\)
= \(\frac{1}{3456000}\)
k mik nha! (kb nhé!!!)
\(\frac{1}{20}\cdot\frac{1}{30}\cdot\frac{1}{42}\cdot\frac{1}{56}\cdot\frac{1}{72}\cdot\frac{1}{90}\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{4}-\frac{1}{10}\)
\(=\frac{6}{40}=\frac{3}{20}\)
1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7+1/7x1/8+1/8x1/9
=7/18
ta co:
6n +11 chia het 2n + 1
=> 6n +11 - 3( 2n +1 ) chia het 2n +1
=> 6n +11 - 6n - 3 chia het 2n +1
=> 8 chia het 2n +1
=> 2n +1 thuoc uoc cua 8
=> 2n +1 thuoc {.......} ban tu liet ke nhe!
=> 2n thuoc { .........} ban tu kiet ke
=> n thuoc {.......} ban tu kiet ke
=> n= -1
tick nha!
<=>3(n+1)+10 chia hết 2n+1
=>30 chia hết 2n+1
=>2n+1\(\in\)U(30)={....} bạn tự liệt kê
=>n\(\in\){....} lấy U(30) chia cho 2 rồi -1
\(3+3^2+.....+3^{99}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{97}+3^{98}+3^{99}\right)\)
\(=39+3^3\left(3+3^2+3^3\right)+........+3^{96}\left(3+3^2+3^3\right)\)
\(=39+3^3\cdot39+...+3^{96}\cdot39\)
\(=39\left(1+3^3+....+3^{96}\right)\)
Vì \(39⋮13\Rightarrow39\in B\left(13\right)\)
\(\overline{ab}+\overline{ba}\)
\(=10a+b+10b+a\)
\(=11a+11b\)
\(=11.\left(a+b\right)⋮11\)
Vậy \(\left(\overline{ab}+\overline{ba}\right)⋮11\)
