S = 3¹⁰⁰ - 1

C = (1 + 3 + 3^2 + 3^3) + ......... + (3^8 + 3^9 + 3^10 + 3^11)

C = 1.40 + .............. + 3^8.40

= 40.(1 + 3^4 + ...... + 3^8)

Chia hết cho 40

A=(3^0+3^1+3^2+3^3)+(3^4+3^5+3^6+3^7)+...+(3^2009+3^2010+3^2011+3^2012)

A=40+3^4*(1+3+3^2+3^3)+...+3^2009*(1+3+3^2+3^3)

A-1=40+80*40+...+3^2009*40

A-1=40*(1+80+..+3^2009)

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)

A= 1 +(3^1+3^2+3^3+3^4)+..............................+(3^2009+3^2010+3^2011+3^2012)

A=1+120+................................+3^2009*(3^1+3^2+3^3+3^4)

A=1+(1+.....................+3^2009)*120

Vì 120 chia hết cho 40

suy ra (1+..........................+3^2009) chia hết cho 40

suy ra A chia 40 dư 1

suy ra A-1 chia hết cho 40

nhưng bạn có tịk ko

có bạn dạy mình đã giải rõ ra cơ