Chứng tỏ rằng 67775-4-1 = 75257-5-2
Cho S = 1+2+2^2+2^3+2^4+2^5+2^6+2^7
Chứng tỏ rằng S chia hết cho 4
Cho S = 1+2+2^2+2^3+2^4+2^5+2^6+2^7
Chứng tỏ rằng S chia hết cho 4 VÀ 13
a) Cho P = 1 + 3 + 32 + 33 +.......+ 3101. Chứng tỏ rằng P⋮13.
b) Cho B = 1 + 22 + 24 +.......+ 22020. Chứng tỏ rằng B ⋮ 21.
c) Cho A = 2 + 22 + 23 +........+ 220. Chứng tỏ A chia hết cho 5.
d) Cho A = 1 + 4 + 42 + 43 +..........+ 498. Chứng tỏ A chia hết cho 21.
e) Cho A = 119 + 118 + 117 +.........+ 11 + 1. Chứng tỏ A chia hết cho 5.
a) P = 1 + 3 + 3² + ... + 3¹⁰¹
= (1 + 3 + 3²) + (3³ + 3⁴ + 3⁵) + ... + (3⁹⁹ + 3¹⁰⁰ + 3¹⁰¹)
= 13 + 3³.(1 + 3 + 3²) + ... + 3⁹⁹.(1 + 3 + 3²)
= 13 + 3³.13 + ... + 3⁹⁹.13
= 13.(1 + 3³ + ... + 3⁹⁹) ⋮ 13
Vậy P ⋮ 13
b) B = 1 + 2² + 2⁴ + ... + 2²⁰²⁰
= (1 + 2² + 2⁴) + (2⁶ + 2⁸ + 2¹⁰) + ... + (2²⁰¹⁶ + 2²⁰¹⁸ + 2²⁰²⁰)
= 21 + 2⁶.(1 + 2² + 2⁴) + ... + 2²⁰¹⁶.(1 + 2² + 2⁴)
= 21 + 2⁶.21 + ... + 2²⁰¹⁶.21
= 21.(1 + 2⁶ + ... + 2²⁰¹⁶) ⋮ 21
Vậy B ⋮ 21
c) A = 2 + 2² + 2³ + ... + 2²⁰
= (2 + 2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷ + 2⁸) + ... + (2¹⁷ + 2¹⁸ + 2¹⁹ + 2²⁰)
= 30 + 2⁴.(2 + 2² + 2³ + 2⁴) + ... + 2¹⁶.(2 + 2² + 2³ + 2⁴)
= 30 + 2⁴.30 + ... + 2¹⁶.30
= 30.(1 + 2⁴ + ... + 2¹⁶)
= 5.6.(1 + 2⁴ + ... + 2¹⁶) ⋮ 5
Vậy A ⋮ 5
d) A = 1 + 4 + 4² + ... + 4⁹⁸
= (1 + 4 + 4²) + (4³ + 4⁴ + 4⁵) + ... + (4⁹⁷ + 4⁹⁸ + 4⁹⁹)
= 21 + 4³.(1 + 4 + 4²) + ... + 4⁹⁷.(1 + 4 + 4²)
= 21 + 4³.21 + ... + 4⁹⁷.21
= 21.(1 + 4³ + ... + 4⁹⁷) ⋮ 21
Vậy A ⋮ 21
e) A = 11⁹ + 11⁸ + 11⁷ + ... + 11 + 1
= (11⁹ + 11⁸ + 11⁷ + 11⁶ + 11⁵) + (11⁴ + 11³ + 11² + 11 + 1)
= 11⁵.(11⁴ + 11³ + 11² + 11 + 1) + 16105
= 11⁵.16105 + 16105
= 16105.(11⁵ + 1)
= 5.3221.(11⁵ + 1) ⋮ 5
Vậy A ⋮ 5
chứng tỏ rằng:
1/4<1/5+2/5^2+3/5^3+...+2016/5^2016<1/3
Chứng tỏ rằng :
\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\) > \(\dfrac{4}{5}\)
Ta có :
\(\dfrac{1}{2}>\dfrac{1}{5}\)
\(\dfrac{1}{3}>\dfrac{1}{5}\)
\(\dfrac{1}{4}>\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\)
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{7}{6}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{17}{12}+\dfrac{1}{5}\)
\(=\dfrac{97}{60}\)
\(\dfrac{4}{5}=\dfrac{4.12}{5.12}=\dfrac{48}{60}\)
Mà \(\dfrac{97}{60}>\dfrac{48}{60}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).
Giúp Mình mấy bài này với nhe!!!
1. Cho Y = 1+3+32+33+.....+398
Chứng tỏ rằng Y⋮13.
2. Cho A = 1+3+32+33.....+32018+32019
Chứng tỏ rằng A⋮4.
3. 2.(x+4)+5=65 (Tìm x).
4.Cho A = 119+ 118+117+.....+11+1. Chứng minh rằng A⋮5. Phần A nha!!!
B) Chứng minh rằng với mọi số tự nhiên n thì n2+n+1 không chia hết cho 4.
5. a) 96-3.(x+1)=42 ( Tìmx )
b) 15x-9x+2x=72
c) 3x+2+3x=10
6. a) 125-3.(x+8)=77
b) (7x-11)3= 22.52- 73
c) 5x+1+5x+2= 750
d) (2x-1)2018= (2x-1)2019.
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
bài 7 chứng tỏ rằng 2/5 < 1/2²+1/3²+1/4²+...+1/100²<1
Chứng tỏ rằng: 1/2*3+1/3*4+1/4*5+....+1/99*100<1/2
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}< \frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}< \frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\left(đpcm\right)\)
Bài 1:
a) Chứng tỏ rằng : 200 - (3+2/3+2/4+....+2/100)
--------------------------------------- = 2
1/2+2/3+3/4+....+9/100
b) Cho B =5/2.1 + 4/1.11 + 3/11.2 + 1/2.15 + 15/4.43 + 13/43
Chứng tỏ rằng B > 3
Cho s=1/1×2+1/2×3+1/3×4+1/4×5+...1/2019×2020 chứng tỏ rằng s
Chứng tỏ rằng : \(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}< \dfrac{4}{9}\)
Đặt \(A=\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{60^2}\)
\(A< \dfrac{1}{3^2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{59.60}\)
\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{59}-\dfrac{1}{60}\)
\(A< \dfrac{1}{3^2}+\dfrac{1}{3}-\dfrac{1}{60}\)
\(A< \dfrac{4}{9}-\dfrac{1}{60}< \dfrac{4}{9}\) (đpcm)