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\(S=2^0+2^1+2^2+...+2^{99}+2^{100}\)
\(=1+2+\left(2^2+2^3+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=3+2^2.\left(1+2+4\right)+...+2^{98}.\left(1+2+4\right)\)
\(=3+7.\left(2^2+2^5+...+2^{98}\right)\)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
\(A=1+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+...+2^{2019}\)
\(2A-A=\left(2+2^2+...+2^{2019}\right)-\left(1+2^2+2^3+...+2^{2018}\right)\)
\(A=2^{2019}-1\)
\(\Rightarrow A+1=2^{2019}-1+1=2^{2019}\)
\(\Rightarrow A+1\)là một lũy thừa
đpcm
a) 3x.3 = 243
3x + 1 = 35
x + 1 = 5
x = 4
b) x2 = x
x2 - x = 0
x(x - 1) = 0
x = 0 hoặc x - 1 = 0
x = 1
d) 64.4x = 16
4x = 16 : 64
4x = 1/4
4x = 4-1
x = -1
a) 4x + 32 = 3.25
4x =64
X =3
b) 86 - 5( x+8) =36
5(x+8)= 50
x= 2
c) 38-3./x/ -5.(24-22.3) { đề lỗi r hay s đấy bn }
d) 2018 < /x/ < 2020
X= - 2019, -2020, 2019, 2020
\(\left[\left(3x+1\right)^3\right]^5=15^0\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1\)
\(\Leftrightarrow\left(3x+1\right)^{15}=1^{15}\)
\(\Rightarrow3x+1=1\)
\(\Leftrightarrow3x=1-1\)
\(\Leftrightarrow3x=0\Rightarrow x=0\)
\(\left[(3\times+1)^3\right]^5=15^0\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1\)
\(\Rightarrow\left[(3\times+1)^3\right]^5=1^5\)
\(\Rightarrow(3\times+1)^3=1\)
\(\Rightarrow(3\times+1)^3=1^3\)
\(\Rightarrow3\times+1=1\)
\(\Rightarrow3\times=1-1\)
\(\Rightarrow3\times=0\)
\(\Rightarrow\times=0\)
a) 5^x=5^78:5^14(lấy 78-14)
5^x=5^64
=> x=64
b) 7^x.7^2=7^21
7^x=7^21:7^2
7^x=7^19
=> x=19
64.4x=16864.4x=168
82.4x=16882.4x=168
4x=168:824x=168:82
4x=4134x=413
⇒x=13⇒x=13
Vậy x=13
64 . 4x = 168
<=> 26 . 22x = 232
<=> 22x = 226
<=> 2x = 26
<=> x = 13
Xin k ạ!
HT