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Ta thấy : các số hạng trong tổng S đều \(>\frac{7}{35}\)
\(\Rightarrow S>\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}+\frac{7}{35}\)
\(\Rightarrow S>\frac{35}{35}\)
\(\Rightarrow S>1\) ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: \(\frac{31+32+35}{34}=\frac{31}{34}+\frac{32}{34}+\frac{35}{34}.\)
mà \(\frac{31}{32}>\frac{31}{34};\frac{32}{33}>\frac{32}{34}\)
\(\Rightarrow\frac{31}{32}+\frac{32}{33}+\frac{35}{34}>\frac{31}{34}+\frac{32}{34}+\frac{35}{34}=\frac{31+32+35}{34}\)
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a,1-3+5-7+9-.......+33-35
=(1+5+9+....+33)-(3+7+11+...+35)
=153-171
=-18
Tick mk vài cái lên 300 mk giải nốt phần b
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a, 6100 - 1 = (6 . 6 . 6 ..... 6) - 1 = [(...6) . (...6) . (...6) ..... (...6)] - 1 = (...6) - 1 = ...5 \(⋮\) 5
b, 2120 - 1110 = (21 . 21 . 21 . 21 . 21..... 21) - (11 . 11 . 11 . 11 ..... 11) = [(...1) . (...1) . (...1) . (...1).....(...1)] - [(...1) . (...1) . (...1) . (...1).....(...1)] = (...1) - (...1) = ....0 \(⋮\) 2; \(⋮\) 5
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\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
M=13+232+...+100311M=13+232+...+1003100
⇒3M=1+23+332+...+10310⇒3M=1+23+332+...+100399
⇒3M−M=(1+23+332+...+10310)−(13+232+...+10310)⇒3M−M=(1+23+332+...+100399)−(13+232+...+1003100)
⇒2M=1+(13+132+...+1399)−1003100⇒2M=1+(13+132+...+1399)−1003100
⇒2M=1+12−1399.2−1003100⇒2M=1+12−1399.2−1003100
⇒M=34−1399.4−503100