a. Ta có: a > b

4a > 4b ( nhân cả 2 vế cho 4)

4a - 3 > 4b - 3 (cộng cả 2 vế cho -3)

b. Ta có: a > b

-2a < -2b ( nhân cả 2 vế cho -2)

1 - 2a < 1 - 2b (cộng cả 2 vế cho 1)

d. Ta có: a < b 

-2a > -2b ( nhân cả 2 vế cho -2)

5 - 2a > 5 - 2b (cộng cả 2 vế cho 5)

\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}=\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\right)>\left(\frac{1}{11}+\frac{1}{11}+...+\frac{1}{11}\right)+\left(\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\right)=\frac{8}{11}+\frac{8}{19}=\frac{240}{209}>\frac{209}{209}=1\Rightarrow B>1\)

a: \(P=5+5^2+5^3+5^4+\cdots+5^{102}\)

\(=\left(5+5^2\right)+\left(5^3+5^4\right)+\cdots+\left(5^{101}+5^{102}\right)\)

\(=5\left(1+5\right)+5^3\left(1+5\right)+\cdots+5^{101}\left(1+5\right)\)

\(=6\left(5+5^3+\cdots+5^{101}\right)\) ⋮6

b:Sửa đề: \(A=1+4+4^2+4^3+\cdots+4^{99}\)

\(=\left(1+4\right)+\left(4^2+4^3\right)+\cdots+\left(4^{98}+4^{99}\right)\)

\(=\left(1+4\right)+4^2\left(1+4\right)+\cdots+4^{98}\left(1+4\right)\)

\(=5\left(1+4^2+\cdots+4^{98}\right)\) ⋮5

c: \(B=1+2+2^2+\cdots+2^{98}\)

\(=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+\cdots+\left(2^{96}+2^{97}+2^{98}\right)\)

\(=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+\cdots+2^{96}\left(1+2+2^2\right)\)

\(=7\left(1+2^3+\cdots+2^{96}\right)\) ⋮7

d:Sửa đề: \(C=1+3+3^2+3^3+\cdots+3^{103}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\cdots+\left(3^{100}+3^{101}+3^{102}+3^{103}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+\cdots+3^{100}\left(1+3+3^2+3^3\right)\)

\(=40\left(1+3^4+\cdots+3^{100}\right)\) ⋮40

\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)

\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}\right)\)

\(\text{Mà:}\)\(\frac{1}{5}+\frac{1}{6}+...+1>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{5}{9}>\frac{1}{2}\)

\(\text{Mà:}\)\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{10}{19}>\frac{1}{2}\)

\(\text{Vậy:}\)\(B>1\)

20A=20+10+40/6+5+4+20/6

=20+10+5+4+(40/6+20/6)

=39+10

=49 chia hết cho 7

1. a là số tự nhiên chia 5 dư 1

=> a = 5k + 1 ( k thuộc N )

b là số tự nhiên chia 5 dư 4

=> b = 5k + 4 ( k thuộc N )

Ta có ( b - a )( b + a ) = b² - a²

                                   = ( 5k + 4 )² - ( 5k + 1 )²

                                   = 25k² + 40k + 16 - ( 25k² + 10k + 1 )

                                   = 25k² + 40k + 16 - 25k² - 10k - 1

                                   = 30k + 15

                                   = 15( 2k + 1 ) chia hết cho 5 ( đpcm )

2. 2n²( n + 1 ) - 2n( n² + n - 3 )

= 2n³ + 2n² - 2n³ - 2n² + 6n

= 6n chia hết cho 6 ∀ n ∈ Z ( đpcm )

3. n( 3 - 2n ) - ( n - 1 )( 1 + 4n ) - 1

= 3n - 2n² - ( 4n² - 3n - 1 ) - 1

= 3n - 2n² - 4n² + 3n + 1 - 1

= -6n² + 6n

= -6n( n - 1 ) chia hết cho 6 ∀ n ∈ Z ( đpcm )