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![](https://rs.olm.vn/images/avt/0.png?1311)
A = 21 + 22 + 23 + ..... + 2100
2A = 22 + 23 +24 + ... + 2100 + 2101
2A - A = A = ( 2101 + 2100 + ... + 22 ) - ( 2100 + 299 + ... + 21 )
A = 2101 - 21
A = 2101 - 2
Hok tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^{2019}-4^{2018}-4^{2017}-...-4-1\)
Đặt \(A=1+4+...+4^{2017}+4^{2018}+4^{2019}\)
\(\Leftrightarrow4A=4+4^2+...+4^{2018}+4^{2019}+4^{2020}\)
\(\Rightarrow4A-A=4^{2020}-1\)
\(\Rightarrow3A=4^{2020}-1\Leftrightarrow A=\frac{4^{2020}-1}{3}\)
\(\Leftrightarrow-A=\frac{1-4^{2020}}{3}\)
Vậy ....
![](https://rs.olm.vn/images/avt/0.png?1311)
(723 . 542) :1084
=373248.2916:136048896
=1088391168:136048896
=8
Chúc bạn học tốt
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\(\left[\left(3x+1\right)^3\right]^5=15^0\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1^{15}\)
\(\Rightarrow3x+1=1\)
\(\Leftrightarrow3x=1-1\)
\(\Leftrightarrow3x=0\Rightarrow x=0\)
\(\left[(3\times+1)^3\right]^5=15^0\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1^5\)
\(\Rightarrow(3\times+1)^3=1\)
\(\Rightarrow(3\times+1)^3=1^3\)
\(\Rightarrow3\times+1=1\)
\(\Rightarrow3\times=1-1\)
\(\Rightarrow3\times=0\)
\(\Rightarrow\times=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2.2^2.2^3.2^4.........2^{100}\)
\(=2^{1+2+3+4+......+100}\)
\(=2^{5050}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=3+2^2.\left(1+2+4\right)+...+2^{98}.\left(1+2+4\right)\)
\(=3+7.\left(2^2+2^5+...+2^{98}\right)\)chia 7 dư 3
\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(S=\left(2^0+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{98}+2^{99}+2^{100}\right)\)
\(S=\left(1+2+4\right)+2^3\left(1+2+4\right)+.....+2^{98}\left(1+2+4\right)\)
\(S=7+2^3\cdot7+....+2^{98}\cdot7\)
\(S=7\left(1+2^3+...+2^{98}\right)\)
=> S chia 7 dư 0 hay S chia hết cho 7
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3^{2x+2}=3^{2\left(x+3\right)}\)
=> 2x + 2 = 2 ( x+ 3 )
=> 2x + 2 = 2x + 6
=> 2x - 2x = 6 - 2
=> 0x = 4 ( loại )
Vậy không có số x thỏa mãn
Ta có:
9x+3=(32)x+3=32x+6=32x+2
=> 2x+6=2x+2 (vô lý)
Vậy ko có số x thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)
Mk nghĩ đề câu 1 là chứng minh 215+211 chia hết cho 17.
Đây là cách giải của mk:
215+211= 211(24+1)= 211(16+1)= 211.17 chia hết cho 17.
=> 215+211 chia hết cho 17.
=3x16-2x9=48-18=30
3.4.4-2.3.3
=6