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Câu 2:
\(C=3^{10}+3^{11}+3^{12}+...+3^{17}.\)
\(C=\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+3^{15}+3^{16}+3^{17}\right).\)
\(C=3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right).\)
\(C=3^{10}\left(1+3+9+27\right)+3^{14}\left(1+3+9+27\right).\)
\(C=3^{10}.40+3^{14}.40.\)
\(C=\left(3^{10}+3^{14}\right).40⋮40\left(đpcm\right).\)
\(C=3^{10}+3^{11}+..+3^{17}\\ =\left(3^{10}+3^{11}+3^{12}+3^{13}\right)+\left(3^{14}+..+3^{17}\right)\\ =3^{10}\left(1+3+3^2+3^3\right)+3^{14}\left(1+3+3^2+3^3\right)\\ =40\left(3^{10}+3^{14}\right)⋮40\)
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a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
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1+2+3+4+5+6+7+8+9+10=55
11+12+13+14+15+16+17+18+19+20=155
1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1+2+3+4+5+6+7+8+9+10=55\)
\(11+12+13+14+15+16+17+18+19+20=155\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
1/2<1/1.2
1/3<1/2.3
...
1/32<1/31.32
=>1/2+1/3+...+1/32<1/1.2+1/2.3+...+1/31.32
=>1/2+1/3+...+1/32<1/1-1/2+1/2-1/3+...+1/31-1/32
=>1/2+1/3+...+1/32<1/1-1/32=31/32
vì 31/32<1
=>tổng đó <1
ta lại có 1+1=2 mà 2 <3
=>tổng đó <3
vậy:-------(bn tự lm nha)
k cho mik vs nha
![](https://rs.olm.vn/images/avt/0.png?1311)
a) + Ta có: \(10^{25}+5=100...0\) ( 25 số 0 ) \(+5=100...05\)( 24 số 0 )
Ta lại có: \(1+0+0+...+0+5=6⋮3\)
\(\Rightarrow10^{25}+5⋮3\)
+ Ta có: \(\hept{\begin{cases}10^{25}⋮5\\5⋮5\end{cases}}\)\(\Rightarrow\)\(10^{25}+5⋮5\)
Vậy \(10^{25}+5⋮3\)và \(5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bn tham khảo câu trả lời của mik ở bn hoàng thu trang nhabây h mik ghi lại dài dòng lắm
\(A=\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{9}\right)+...+\left(1+\frac{1}{900}\right).\)
\(=29+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{30^2}\)
\(< 29+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{29.30}\)
\(< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{29}-\frac{1}{30}< 29+1=30\)