Biến đổi vế phải ta có \(\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\Leftrightarrow a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)

\(\Leftrightarrow a^5+b^5+\left(a+b\right)-\left(a+b\right)=a^5+b^5\) (vì ab=1)

1) C = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰  

       = (5 + 5²) + (5³ + 5⁴) + ... +(5¹⁹ + 5²⁰)

       = (5 + 5²) + 5²(5 + 5²) + .... + 5¹⁸(5 + 5²) 

       = (5 + 5²)(1 + 5² + ... + 5¹⁸)

       = 26(1 + 5² + ... + 5¹⁸)

        = 13.2.(1 + 5² + ... + 5¹⁸) \(⋮\)13 (ĐPCM)

2) a) A = 2⁴ + 2⁵ + 2⁶ + 2⁷ + 2⁸ + 2⁹ 

           = (2⁴ + 2⁵) + (2⁶ + 2⁷) + (2⁸ + 2⁹)

           = 2⁴(1 + 2) + 2⁶(1 + 2) + 2⁸(1 + 2)

           = (1 + 2)(2⁴ + 2⁶ + 2⁸)

           = 3(2⁴ + 2⁶ + 2⁸) \(⋮3\)

b) B = 3¹⁷ + 3¹⁸ + 3¹⁹ + 3²⁰ + 3²¹ + 3²² 

      = (3¹⁷ + 3¹⁸ + 3¹⁹) + (3²⁰) + 3²¹ + 3²²) 

      = 3¹⁷(1 + 3 + 3²) + 3²⁰(1 + 3 + 3²)

      = (1 + 3 + 3²)(3¹⁷ + 3²⁰)

      = 13(3¹⁷ + 3²⁰) \(⋮\)13

Bài 1:

C = 5+5² +5³+.....+5²⁰

=(5+5²+5³+5⁴)+.....+(5¹⁷+5¹⁸+5¹⁹+5²⁰)

=5.(1+5+5²+5³)+.....+5¹⁷(1+5+5²+5³)

=5.156+....+5¹⁷.156

=156.(5+...+5¹⁷)=13.12.(5+....+5¹⁷) chia hết cho 13

Bài 2:

A=2⁴+2⁵+2⁶+2⁷+2⁸+2⁹

=(2⁴+2⁵)+(2⁶+2⁷)+(2⁸+2⁹)

=2⁴(1+2)+2⁶(1+2)+2⁸(1+2)

=2⁴.3+2⁶.3+2⁸.3

=3.(2⁴+2⁶+2⁸) chia hết cho 3 

b)

B=3¹⁷+3¹⁸+3¹⁹+3²⁰+3²¹+3²²

=(3¹⁷+3¹⁸+3¹⁹)+(3²⁰+3²¹+3²²)

=3¹⁷(1+3+3²)+3²⁰(1+3+3²)

=3¹⁷.13+3²⁰.13

=13.(3¹⁷+3²⁰)chia hết cho 13

#CừU

Bài 3: 

a: \(n\left(2n-3\right)-2n\left(n+1\right)\)

\(=2n^2-3n-2n^2-2n\)

=-5n chia hết cho 5

b: \(\left(n-1\right)\left(n+4\right)-\left(n-4\right)\left(n+1\right)\)

\(=n^2+4n-n-4-\left(n^2+n-4n-4\right)\)

\(=n^2+3n-4-\left(n^2-3n-4\right)\)

\(=6n⋮6\)

B2

( a³ + a²b + ab² + b³ ).( a - b ) = a⁴ - b⁴

[( a³ + b³ + ab.( a + b )].( a - b ) = a⁴ - b⁴

[( a + b ).( a² - ab + b² ) + ab.( a + b )].( a - b ) = a⁴ - b⁴

 ( a + b ).( a² - ab + b² + ab ).( a - b ) = a⁴ - b⁴

( a + b ).( a² + b² ).( a  -  b ) = a⁴ - b⁴

 ( a² - b² ).( a² + b² ) = a⁴ - b⁴

 a⁴ - b⁴ = a⁴ - b⁴  ( đpcm )

1: \(\Leftrightarrow a^5-a^4b+b^5-ab^4>=0\)

\(\Leftrightarrow a^4\left(a-b\right)-b^4\left(a-b\right)>=0\)

\(\Leftrightarrow\left(a-b\right)^2\cdot\left(a+b\right)\cdot\left(a^2+b^2\right)>=0\)(luôn đúng khi a,b dương)

\(\dfrac{a}{a+2\sqrt{\left(a+bc\right)}}=\dfrac{a}{a+2\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{a}{a+2\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(=\dfrac{a}{a+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}+\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\)

\(\le\dfrac{a}{5^2}\left(\dfrac{1}{a}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}+\dfrac{1}{\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{2}}\right)\)

\(=\dfrac{a}{25}\left(\dfrac{1}{a}+\dfrac{8}{\sqrt{\left(a+b\right)\left(a+c\right)}}\right)=\dfrac{1}{25}+\dfrac{8}{25}.\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)

Tương tự:

\(\dfrac{b}{b+2\sqrt{b+ac}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\)

\(\dfrac{c}{c+2\sqrt{c+ab}}\le\dfrac{1}{25}+\dfrac{4}{25}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)

Cộng vế:

\(P\le\dfrac{3}{25}+\dfrac{4}{25}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{15}{25}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)