Cách đây nè:

  2^10 = 1025 ; 7^3 = 343 => 2^10 < 3.7^3 => \(\left(2^{10}\right)^{238}<3^{238}.\left(7^3\right)^{238}\)  => 2^2380 < 3^238 .7^714

2^8 = 256 ; 3^5 = 243 => 3^5 < 2^8

Ta có :

  3^328 = 3^3 . 3^325 = \(3^3.\left(3^5\right)^{47}<3^3.\left(2^8\right)^{47}<2^5.2^{376}\Rightarrow3^{328}<2^{381}\)

2^2380 <  2^238 . 7^714  => 2^2380 < 2^238 . 7^714 => 2^1999 < 7^14 mà 2^1999 > 2^1993 => 2^1993 < 7^714

Chỉnh lại tý

2^10 = 1025 ; 7^3 = 343 => 2^10 < 3.7^3 => \(\left(2^{10}\right)^{238}<3^{238}.\left(7^3\right)^{238}\)  => 2^2380 < 3^238 .7^714

2^8 = 256 ; 3^5 = 243 => 3^5 < 2^8

Ta có :

  3^328 = 3^3 . 3^325 = \(3^3.\left(3^5\right)^{47}<3^3.\left(2^8\right)^{47}<2^5.2^{376}\Rightarrow3^{328}<2^{381}\)

 2^2380 < 2^238 . 7^714 => 2^1999 < 7^14 mà 2^1999 > 2^1993 => 2^1993 < 7^714

J

hoc moi lop 5 

a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004

             B=    1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005

suy ra 2B=1-1/3^2005

    suy ra B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)

suy ra B=1/2-1/3^2005/2 bé hơn 1/2

từ đấy suy ra B bé hơn 1/2

\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)

\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)

\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)

\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)

\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)

\(B=1-\frac{1}{100}< 1\)

=> B < 1 (Đpcm)

B = 3/1².2² + 5/2².3² + 7/3².4² + ... + 19/9².10²

B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100

B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100

B = 1 - 1/100 < 1 ( đpcm)

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

=> đpcm

Ủng hộ mk nha ^_-

đặt \(S=1+4+4^2+......+4^{1999}\)

\(\Rightarrow4S=4+4^2+4^3+....+4^{2000}\)

\(\Rightarrow4S-S=\left(4+4^2+4^3+....+4^{2000}\right)-\left(1+4+4^2+.....+4^{1999}\right)\)

\(\Rightarrow3S=4^{2000}-1\Rightarrow S=\frac{4^{2000}-1}{3}\)

Khi đó \(A=75.S=75.\frac{4^{2000}-1}{3}=\frac{75.\left(4^{2000}-1\right)}{3}=\frac{75}{3}.\left(4^{2000}-1\right)=25.\left(4^{2000}-1\right)=25.4^{2000}-25\)

Ta có: 4²⁰⁰⁰-1=(4⁴)⁵⁰⁰-1=(...6)-1=....5

=>25.4²⁰⁰⁰-25=25.(....5)-25=(...5)-25=....0 chia hết cho 100

Vậy ta có điều phải chứng minh
 

75 chia hết cho 25.

4²⁰⁰⁷ + ... + 4 + 1 chia 4 dư 1 hay không chia hết cho 4

=> 75(4²⁰⁰⁷ + ... + 4 + 1) không chia hết cho 100.

\(\frac{a}{c}=\frac{c}{b}\Rightarrow\frac{a^2}{c^2}=\frac{c^2}{b^2}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{a^2}{c^2}=\frac{c^2}{b^2}=\text{​​}\frac{a^2+c^2}{c^2+b^2}=\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{c}{b}=\frac{a}{b}\)

=> \(\frac{a}{b}=\frac{a^2+c^2}{b^2+c^2}\left(đpcm\right)\)

b) \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55⋮55\left(đpcm\right)\)

a) Từ \(\frac{a}{c}=\frac{c}{b}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{c}{b}\right)^2=\frac{a^2}{c^2}=\frac{c^2}{b^2}=\frac{a^2+c^2}{c^2+b^2}\)(1)

Ta có \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{c}{b}=\frac{a}{b}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{a^2+c^2}{c^2+b^2}=\frac{a}{b}=\left(\frac{a}{c}\right)^2\left(đpcm\right)\)

b) Ta có \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.55⋮55\left(đpcm\right)\)