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b) \(A=3^1+3^2+3^3+...+3^{2006}\)
\(=3+3^2+\left(3^3+3^4+3^5+3^6\right)+....+\left(3^{2003}+3^{2004}+3^{2005}+3^{2006}\right)\)
\(=12+3^3\left(1+3+3^2+3^3\right)+...+3^{2003}\left(1+3+3^2+3^3\right)\)
\(=12+\left(1+3+3^2+3^3\right)\left(3^3+...+3^{2003}\right)\)
\(=12+40\left(3^3+...+3^{2003}\right)\)
\(=12+.....0=.....2\)
Vậy A có tận cùng là chữ số 2
Ta có: 2^30= 2^3.10= (2^3)^10= 8^10
3^20= 3^2.10= (3^2)^10= 9^10
Vì 8<9 nên 8^10<9^10
=> 2^30<3^20
a) 12 . x - 33 = 243
12 . x = 243 + 33
12 . x = 276
x = 276 : 12
x = 23
a, \(12x-33=3^2.3^3\)
\(\Leftrightarrow12x-33=3^{2+3}=3^5\)
\(\Leftrightarrow12x-33=243\Rightarrow12x=243+33=276.\)
\(\Leftrightarrow12x=276\Leftrightarrow x=276:12\)
\(\Rightarrow x=23\)
b) \(7x-3^3=2^7:2^4\)
\(\Leftrightarrow7x-27=2^{7-4}=2^3=8\)
\(\Leftrightarrow7x=8+27=35\)
\(\Leftrightarrow x=35:7\)
\(\Leftrightarrow x=5\)
c) \(3^x.3=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Rightarrow x+1=5\Rightarrow x=5-1\)
\(\Rightarrow x=4\)
d) \(2^x.7=56\Rightarrow2^x=56:7=8\)
\(\Rightarrow2^x=2^3\Rightarrow x=3\)
20175 x ( x - 10 ) = 20176
Vì 20176 = 2017 x 2017 x 2017 x.....x 2017
Nên => x - 10 phải = 2017
=> x = 2017 + 10 = 2027
(x5)10 = x3
x chỉ có thể = 1 vì x5 > x3
=> x = 1
2017^5.(x-10)=2017^6
x-10=2017^6/2017^5
x-10=2017
x=2017+10
x=2027
(x^5)^10=x^3
x^50=x^3
x^50-x^3=0
x^48.x^3-x^3.1=0
x^3.(x^48-1)=0
<=>[x^3=0
x^48-1=0
<=>x^3=0^3
x^48=0+1
<=>x=0
x^48=1
<=>x=0
x648=1^48
<=>x=0
x=1
vay x=0 hoac x=1
x^2005=x^2
x^2005-x^2=0
x^2003*x^2*1=0
x^2.(x^2003-1)
x^2=0
x^2003-1=0
x^2=0^2
x^2003=0+1
x=0
x^2003=1
x=0
x^2003=1^2003
x=0
x=1
vay x=0 hoac x=1
\(3^4.3^x:9=3^7\)
\(\Leftrightarrow3^{4+x}:3^2=3^7\)
\(\Leftrightarrow3^{4+x-2}=3^7\)
\(\Rightarrow2+x=7\)
\(\Leftrightarrow x=5\)
Ta có
x ≥ 0 √ x ∈ Z
=> x2 + 1 ≥ 1
=> (x2 + 1)2 ≥ 12 = 1
=> F = (x2 + 1)2 + 4 ≥ 1 + 4 = 5
=> F = (x2 + 1)2 + 4 ≥ 5
Dấu "=" xảy ra khi x2 = 0 => x = 2
Vậy GTNN của F là 5 tại x = 0
Chỗ kia mình ấn nhầm ra bạn
Dấu "=" xảy ra khi x2 = 0 => x = 0
cho \(M=1+3+3^2+...+3^{99}+3^{100}\)
=>\(M=1+\left(3+3^2+3^3\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=>M=1+3\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=>M=1+13\left(3+...+3^{98}\right)\)
Mà \(13\left(3+3^{98}\right)⋮13\)
=> M chia cho 13 dư 1
+) \(M=1+3+3^2+...+3^{99}+3^{100}\)
\(\Leftrightarrow M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(\Leftrightarrow M=\left(1+3+9\right)+3^3\left(1+3+9\right)+....+3^{98}\left(1+3+9\right)\)
\(\Leftrightarrow M=13+3^3\cdot14+....+3^{98}\cdot14\)
\(\Leftrightarrow M=13\left(1+3^3+....+3^{98}\right)\)
=> M chia 13 dư 0
12 . ( x - 1 ) : 3 - 43 + 23
= 12 . ( x - 1 ) : 3 - 64 + 8
= 12 . ( x - 1 ) : 3 - 72
= 12 . ( x - 1 ) 72 . 3
= 12 . ( x - 1 ) 216
= ( x - 1 ) 216 : 12
= ( x - 1 ) 18
= x 18 + 1
= x 9
Mình nhầm nhé = 19