S = 1 + 2 + 3 + ..... + 200
A = 1 .2 + 2.3 + .....+ 199.200
B = 11 + 21 + 31 + ..... + 2002
1. CMR: ∀ n∈\(N^{\cdot}\)
a) \(A=5^n+2.3^{n-1}+1\text{⋮}8\)
b) \(B=3^{n+2}+4^{2n+1}\text{⋮}13\)
c) \(C=6^{2n}+3^{n+2}+3^n\text{⋮}11\)
d) \(D=1^n+2^n+5^n+8^n\text{⋮}8\)
2. \(CMR:\) \(1^{2002}+2^{2002}+...+2002^{2002}\text{⋮}11\)
3. a) cho a,b ∈Z, t/m:\(a^2+b^2\text{⋮}7\). \(CMR:a\text{⋮}7;b\text{⋮}7\)
b) \(CMR:\) Nếu \(a^2+b^2\text{⋮}21\) thì \(a^2+b^2\text{⋮}441\) (a,b ∈Z)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
\(2,\)
Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)
Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)
Tính nhanh :
a) S = 991 - 981 + 971 - 961 +.....+ 31 - 21 + 11 - 1
b) S = 1 + 3 + 5 + 7 +......+ (2n - 3) + (2n - 1)
c) 1 + 11 + 21 + 31 +......+ 981 + 991
d) S = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 +......+ 994 - 995 -996 + 997 + 998 (Đ/S : 999)
Bài 1.Tính bằng cách hớp lí
a,(1/4+-5/13)+(2/11+-8/13+3/4)
b,(21/31+-16/7)+(44/53+10/31)+9/53
c,-5/7+3/4+-1/5+-2/7+1/4
d,-3/27+-6/17+1/25+-28/31+-11/17+-1/5
Bài 2.Tính tổng
A=1/1.2+1/2.3+1/3.4+......+1/2003.2004
AI LÀM NHANH VÀ ĐÚNG MK TICK NHA
Bài 1: Ta chỉ cần bỏ ngoặc rồi cộng hai phân số để ra kết quả là số tự nhiên là xong
Bài 2:
A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{2003.2004}\)
A = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-.............-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
A = \(1-\frac{1}{2004}\)
A = \(\frac{2004}{2004}-\frac{1}{2004}=\frac{2003}{2004}\)
Mk làm mẫu 1 bài nha
Bài 1 :
a, = (1/4+3/4) - (5/13+8/13)+2/11
= 1 - 1 + 2/11
= 2/11
Tk mk nha
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2003.2004}\)
\(\Leftrightarrow A=1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\right)\)
\(\Leftrightarrow A=1\left(1-\frac{1}{2004}\right)\)
\(\Leftrightarrow A=\frac{2003}{2004}\)
Chứng minh (1.2-1)/21+(2.3-1)/31+...+(99.100-1)/1001 <2
c) {[(20 - 2.3).5] - 2.5} : 2 + (4.5)²
= {(20 - 6).5] - 10} : 2 + 20²
= (14.5 - 10) : 2 + 400
= (70 - 10) : 2 + 400
= 60 : 2 + 400
= 30 + 400
= 430
d) (1² + 2² + 3² + ... + 100²) . (2⁴ - 4²)
= (1² + 2² + 3² + ... + 100²) . (16 - 16)
= (1² + 2² + 3² + ... + 100²) . 0
= 0
Bài 2
a) -2x < 5
2x > 5
x > 5/2
b) [31 - (x + 5)].11 = 121
31 - (x + 5) = 121 : 11
31 - (x + 5) = 11
x + 5 = 31 - 11
x + 5 = 20
x = 20 - 5
x = 15
c) (x + 1)³ = 27
(x + 1)³ = 3³
x + 1 = 3
x = 3 - 1
x = 2
(1-1/2).(1-1/3). ... .(1-1/2022).x=1-1/1.2-1/2.3-...-1/2002*2003
(1 - \(\dfrac{1}{2}\)).(1 - \(\dfrac{1}{3}\))....(1- \(\dfrac{1}{2022}\)).\(x\) = 1 - \(\dfrac{1}{1.2}\) - \(\dfrac{1}{2.3}\)-...-\(\dfrac{1}{2002.2003}\)
(\(\dfrac{2-1}{2}\)).(\(\dfrac{3-1}{3}\))...(\(\dfrac{2022-1}{2022}\)).\(x\) = 1 - (\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2002.2003}\))
\(\dfrac{1}{2}\).\(\dfrac{2}{3}\)...\(\dfrac{2021}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\)+ ... + \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = 1 - (\(\dfrac{1}{1}\) - \(\dfrac{1}{2003}\))
\(\dfrac{1}{2022}\).\(x\) = \(\dfrac{1}{2003}\)
\(x\) = \(\dfrac{1}{2003}\) : \(\dfrac{1}{2022}\)
\(x\) = \(\dfrac{2022}{2003}\)
1) chứng tỏ tổng 41/90+ 31/72+ 21/40+ -11/45+ -1/36 lớn hơn 1
2) Cho S = 1/5 + 1/6 + ..... + 1/16 + 1/17 .chứng tỏ 1<S<2
Bài : Tính hợp lí nếu có thể :
a. 31 . ( -18 ) + 37 . ( -81 ) - 31
b. - 48 + 48 . ( - 78 ) + 48 . ( -21 )
c. (- 5 ) ^2 - 3. (-2)^3 + 4. (-11 )
d. 13. ( 23 + 22 ) - 3. ( 17 + 28)
e. 1 - 2 - 3 + 4 +5 - 6 - 7 + 8 +.... + 21 - 22 - 23 + 24
P/S : Giúp mình với mơn các bạn :<
Thực hiện phép tính:
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
C) ( -2/3 + 3/7 ) : 4/5 + (-1/3 + 4/7 ) : 4/5
D) 5/9 : ( 1/11 -5/22 ) + 5/9 : ( 1/15 - 2/3 )
E) 3/14 : 1/28 -13/21 : 1/28 + 29/42 : 1/28 - 8
F) ( -40/51 x 0,32 x17/20) : 64/75
Đây là dấu x ( là nhân ) mọi người nhé
mong mn giúp em ạ
A) 7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11
=7/38.(9/11+4/11-2/11)
=7/38
B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20
=5/31.(21/25-7/10-9/20)
=5/31.(-31/100)
=-1/20
1/Chứng minh : 5^2003+5^2002+5^2001 chia hết cho 31'
2./ Cho A = 1+2+2^2+......+2^9+2^10 và B = 2^11- 1 .So sánh A B
2/
A=1+2+2^2+...+2^10
2.A= 2+2^2+...+2^11
=>2A-A = 2^11-1=> A = 2^11 -1=B
Vậy A=B
1)52003+52002+52001=52001(52+5+1)=52001(25+5+1)=52001.31
Vì 31 chia hết cho 31nên
52001.31chia hết cho 31 hay 52003+52002+52001 chia hết cho 31
2) A = 1+2+22+......+29+210
=>2A=2+22+23+...+211
=>2A-A=2+22+23+...+211-(1+2+22+...+29+210)
=>A=211-1
Vậy A=B=211-1