A = 1 + 2 + 2² + ... + 2²⁰⁰⁹ + 2²⁰¹⁰

= ( 1 + 2 + 2² ) + ( 2³ + 2⁴ + 2⁵ ) + ... + ( 2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰ )

= 7 + 2³( 1 + 2 + 2² ) + ... + 2²⁰⁰⁸( 1 + 2 + 2² )

= 7 + 2³.7 + ... + 2²⁰⁰⁸.7

= 7( 1 + 2³ + ... + 2²⁰⁰⁸ ) chia hết cho 7

hay A chia 7 dư 0

Ta có A  = 1 + 2 + 2² + ... + 2²⁰⁰⁹ + 2²⁰¹⁰

=> A - 3 = 2² + 2³ + 2⁴ + 2⁵ + 2⁶ + 2⁷ + .... +  2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰

             = (2² + 2³ + 2⁴) + (2⁵ + 2⁶ + 2⁷) + .... +  (2²⁰⁰⁸ + 2²⁰⁰⁹ + 2²⁰¹⁰)

             = 2²(1 + 2 + 2²) + 2⁵(1 + 2 + 2²) + ... + 2²⁰⁰⁸(1 + 2 + 2²)

             = (1 + 2 + 2²)(2² + 2⁵ + ... + 2²⁰⁰⁸) 

             = 7(2² + 2⁵ + ... + 2²⁰⁰⁸) \(⋮\)7

Vì \(A-3⋮7\)

=> A : 7 dư 3

Vậy A : 7 dư 3

bđt <=> \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ac}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\)

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

\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{a}+\frac{1}{c}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)(1)

Đặt 1/a = x ; 1/b = y ; 1/c = z

(1) được viết lại thành \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\)

Theo Bunyakovsky dạng phân thức ta có :

\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)

Lại có \(x+y+z\ge3\sqrt{xyz}=3\sqrt{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}=3\sqrt{\frac{1}{abc}}=3\)( Cauchy )

=> \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{y+x}\ge\frac{x+y+z}{2}\ge\frac{3}{2}\)

Hay \(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\ge\frac{3}{2}\)( đpcm )

Đẳng thức xảy ra <=> x = y = z = 1 

hay a = b = c = 1

Ta có : a³ + b³ + c³ = 3abc

=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

=> (a + b)³ - 3ab(a + b) + c³ - 3abc = 0

=> [(a + b)³ + c³] - [(3ab(a + b) + 3abc] = 0

=> (a + b + c)(a² + b² + 2ab - ac - bc + c²) - 3ab(a + b + c) = 0

=> (a + b + c)(a² + b² + c² - ab - ac - bc) = 0

=> a² + b² + c² - ab- ac - bc = 0

=> 2(a² + b² + c² - ab- ac - bc) = 0

=> 2a² + 2b² + 2c² - 2ab - 2ac - 2bc = 0

=> (a² - 2ab + b²) + (b² - 2bc + c²) + (a² - 2ac + c²) = 0

=> (a - b)² + (b - c)² + (a - c)² = 0

=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow a=b=c\)

Khi đó M = \(\frac{a^{2020}+b^{2020}+c^{2020}}{\left(a+b+c\right)^{2020}}=\frac{3.c^{2020}}{\left(3c\right)^{2020}}+\frac{3c^{2020}}{3^{2020}.c^{2020}}=\frac{1}{3^{2019}}\)

a) Gọi ƯCLN(2n + 1;3n + 1) = d

=> \(\hept{\begin{cases}2n+1⋮d\\3n+1⋮d\end{cases}}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+1\right)⋮d\end{cases}}\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+2⋮d\end{cases}}\Rightarrow\left(6n+3\right)-\left(6n+2\right)⋮d\)

=> \(1⋮d\Rightarrow d=1\)

=> 2n + 1 ; 3n + 1 là 2 số nguyên tố cùng nhau

b) Gọi ƯCLN(7n + 10 ; 5n + 7) = d

=> \(\hept{\begin{cases}7n+10⋮d\\5n+7⋮d\end{cases}}\Rightarrow\hept{\begin{cases}5\left(7n+10\right)⋮d\\7\left(5n+7\right)⋮d\end{cases}}\Rightarrow\hept{\begin{cases}35n+50⋮d\\35n+49⋮d\end{cases}}\Rightarrow\left(35n+50\right)-\left(35n+49\right)⋮d\)

=> \(1⋮d\Rightarrow d=1\)

=> 7n + 10 ; 5n + 7 là 2 số nguyên tố cùng nhau

Trả L

4 + 5 - 2 + 4 

= 9 - 2 + 4

= 7 + 4

= 11

Hok tốt

4 + 5 - 2 + 4 = 9 - 2 + 4

                    =7 + 4

                    =11