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Bài 1:
Theo đề ra ta có:
$a-2\vdots 3; a-3\vdots 5$
$a-2-2.3\vdots 3; a-3-5\vdots 5$
$\Rightarrow a-8\vdots 3; a-8\vdots 5$
$\Rightarrow a-8=BC(3,5)$
$\Rightarrow a-8\vdots 15$
$\Rightarrow a=15k+8$ với $k$ tự nhiên.
Mà $a$ chia 11 dư 6
$\Rightarrow a-6\vdots 11$
$\Rightarrow 15k+8-6\vdots 11$
$\Rightarrow 15k+2\vdots 11\Rightarrow 4k+2\vdots 11$
$\Rightarrow 4k+2-22\vdots 11\Rightarrow 4k-20\vdots 11$
$\Rightarrow 4(k-5)\vdots 11\Rightarrow k-5\vdots 11$
$\Rightarrow k=11m+5$
Vậy $a=15k+8=15(11m+5)+8=165m+83$ với $m$ tự nhiên.
Vì $a<500\Rightarrow 165m+83<500\Rightarrow m< 2,52$
$\Rightarrow m=0,1,2$
Nếu $m=0$ thì $a=165.0+83=83$
Nếu $m=1$ thì $a=165.1+83=248$
Nếu $m=2$ thì $a=165.2+83=413$
Bài 2:
$a=BC(60,85,90)$
$\Rightarrow a\vdots BCNN(60,85,90)$
$\Rightarrow a\vdots 3060$
Mà $a<1000$ nên $a=0$
a) \(A=1+2+2^2+2^3+...+2^{60}\)
=>\(2A=2+2^2+2^3+2^4+...+2^{61}\)
=>\(2A-A=\left(2+2^2+2^3+2^4+...+2^{61}\right)-\left(1+2+2^2+2^3+...+2^{60}\right)\)
=>\(A=2^{61}-1\)
b) \(B=1+3+3^2+3^3+...+3^{46}\)
=>\(3B=3+3^2+3^3+3^4+...+3^{47}\)
=>\(3B-B=\left(3+3^2+3^3+3^4+...+3^{47}\right)-\left(1+3+3^2+3^3+...+3^{46}\right)\)
=>\(2A=3^{47}-1\)
=>\(B=\frac{3^{47}-1}{2}\)
c) \(C=1+5^2+5^4+...+5^{200}\)
=>\(5^2C=5^2+5^4+5^6+...+5^{202}\)
=>\(25C=5^2+5^4+5^6+...+5^{202}\)
=>\(25C-C=\left(5^2+5^4+5^6+...+5^{202}\right)-\left(1+5^2+5^4+...+5^{200}\right)\)
=>\(24C=5^{202}-1\)
=>\(C=\frac{5^{202}-1}{24}\)
a) A = \(1+2+2^2+2^3+...+2^{60}\)
2A = \(2.\left(1+2+2^2+2^3+...+2^{60}\right)\)
2A = \(2+2^2+2^3+2^4+...+2^{61}\)
2A - A = \(\left(2+2^2+2^3+2^4+...+2^{61}\right)\)- \(\left(1+2+2^2+2^3+...+2^{60}\right)\)
A = \(2^{61}-1\)
b)B = \(1+3+3^2+3^3+...+3^{46}\)
3B = \(3.\left(1+3+3^2+3^3+...+3^{46}\right)\)
3B = \(3+3^2+3^3+3^4+...+3^{47}\)
3B - B = \(\left(3+3^2+3^3+3^4+...+3^{47}\right)\)- \(\left(1+3+3^2+3^3+...+3^{46}\right)\)
2B = \(3^{47}-1\)
B = \(\left(3^{47}-1\right):2\)
Câu a mk ko hiểu gì nha xl bn nhìu
b)1-2+3-4+...+99-100
=(1-2)+(3-4)+...+(99-100)
=(-1)+(-1)+...+(-1)
=(-1) . 50
=(-50)
c) 5 + 52 + 53 + ...+ 599 + 5100
=(5+52)+(53+54)+....+(599+5100)
=30+52(5+52)+...+598(5+52)
=30.1+52.30+.....+598.30
=30(1+52+...+598) chia hết cho 6
Tìm x
a.( x - 140 ) : 3 = 27
x - 140 = 27 . 3
x - 140 = 81
x = 221
b.14 - 4 ( x + 1 ) = 10
4 ( x + 1 ) = 14 - 10
4 ( x +1) = 4
x + 1 = 1
x = 0
c. 15 ( 7 - x ) = 15
7 - x = 1
x = 6
d.34 ( x - 3 ) = 0
\(\Rightarrow\) 34 = 0 hoặc x - 3 = 0
1. 34 = 0 ( vô lí )
2. x - 3 = 0 \(\Rightarrow\) x = 3
e. 24 + 6 (3 - x ) = 30
6( 3- x ) = 30 - 24
6( 3 - x ) = 6
3 - x = 1
x = 2
f. x3 + 24 = 51
x3 = 51 - 24
x3 = 27
\(\Rightarrow\)x = 3 ; x = -3
g. ( x- 5 )2 - 5 = 44
( x - 5) 2 = 49
\(\Rightarrow\)x - 5 = 7 hoặc x - 5 = -7
1. x - 5 = 7\(\Rightarrow\)x = 12
2. x - 5 = -7 \(\Rightarrow\)x = -2
h. ( x + 1 )3 - 23 = 4
( x + 1 )3 =27
\(\Rightarrow\) x + 1 = 3 hoặc x + 1 = -3
1. x + 1 = 3\(\Rightarrow\)x = 2
2. x + 1 = -3 \(\Rightarrow\)x = -4
Bài 1:
$B=1+3+3^2+3^3+...+3^{100}$
$=1+(3+3^2)+(3^3+3^4)+...+(3^{99}+3^{100})$
$=1+3(1+3)+3^3(1+3)+...+3^{99}(1+3)$
$=1+(1+3)(3+3^3+...+3^{99})=1+4(3+3^3+....+3^{99})$
$\Rightarrow B$ chia 4 dư 1.
Bài 2:
$C=5-5^2+5^3-5^4+...+5^{2023}-5^{2024}$
$5C=5^2-5^3+5^4-5^5+...+5^{2024}-5^{2025}$
$\Rightarrow C+5C=5-5^{2025}$
$6C=5-5^{2025}$
$C=\frac{5-5^{2025}}{6}$