11n = 1 33 1
giúp mik
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\(=\dfrac{20}{21}x\dfrac{21}{22}x\dfrac{22}{23}x...x\dfrac{1999}{2000}\)
\(=\dfrac{20}{2000}=\dfrac{1}{100}\)
=20/21x21/22x22/23x..............x1998/1999x1999/2000
=20x21x22x23x.....................x1998x1999/21x22x23x24x...............x1999x2000
=20/2000
1/100
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(\Leftrightarrow S=1\left(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\right)\)
\(\Leftrightarrow S-S=1+\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(\Leftrightarrow S=1-\frac{1}{60}=\frac{59}{60}\)
1) (-37) + 14 + 26 + 37
= [(-37) + 37] + 14 + 26
= 0 + 40
= 40
2) (-24) + 6 + 10 + 24
= [(-24) + 24] + 6 + 10
= 0 + 16
= 16
3) 15 + 23 + (-25) + (-23)
= 15 + (-25) + [(-23) + 23]
= -10 + 0
= -10
4) 60 + 33 + (-50) + (-33)
= 60 + (-50) + [(-33) + 33]
= 10 + 0
= 10
=00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
\(A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{96}.13\)
\(=13\left(1+3^3+...+3^{96}\right)⋮13\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ A=13\left(1+3^3+...+3^{96}\right)⋮13\)
11n + 2 + 122n + 1 = 121 . 11n + 12 . 144n
=(133 – 12) . 11n + 12 . 144n = 133 . 11n + (144n – 11n) . 12
Ta có: 133 . 11n chia hết 133; 144n – 11n chia hết (144 – 11)
144n – 11n chia hết 133 11n + 2 + 122n + 1 chia hết cho 133
chúc bạn học tốt !!!
\(11^n=1331\)
\(\Rightarrow11^3=1331\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
11ⁿ = 1331
11ⁿ = 11³
n = 3