\(\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\dfrac{x^2+\left(x+1\right)^2+x^2\left(x+1\right)^2}{x^2\left(x+1\right)^2}}=\sqrt{\dfrac{x^2\left(x+1\right)^2+2x^2+2x+1}{x^2\left(x+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(x^2+x\right)^2+2\left(x^2+x\right)+1}{\left(x^2+x\right)^2}}=\sqrt{\dfrac{\left(x^2+x+1\right)^2}{\left(x^2+x\right)^2}}=\dfrac{x^2+x+1}{x^2+x}\)

\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

\(5x^2+5y^2+8xy-2x+2y+2=0\)

\(\Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

Vì \(\left(x+y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+1\right)^2\ge0\)

\(\Rightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2\ge0\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

\(\left(x+y\right)^{2018}+\left(x-2\right)^{2019}+\left(y+1\right)^{2020}=\left(1-1\right)^{2018}+\left(1-2\right)^{2019}+\left(-1+1\right)^{2020}=-1\)

1.

Giả sử $a^{2016}+b^{2017}+c^{2018}\vdots 6.$

Ta cần chứng minh $a^{2018}+b^{2019}+c^{2020}\vdots 6.$

Ta có $a^{2018}-a^{2016}=a^{2016}(a^2-1)=a^{2016}(a-1)(a+1).$

Trong ba số nguyên liên tiếp $a-1,\ a,\ a+1$ luôn có một số chia hết cho $3$ và có ít nhất một số chẵn.

Do đó $a(a-1)(a+1)\vdots6.$

Suy ra $a^{2016}(a^2-1)=a^{2015}\cdot a(a-1)(a+1)\vdots6,$ hay $a^{2018}\equiv a^{2016}\pmod6.$

Tương tự, $b^{2019}-b^{2017}=b^{2017}(b^2-1)=b^{2016}\cdot b(b-1)(b+1)\vdots6,$ nên $b^{2019}\equiv b^{2017}\pmod6.$

Lại có $c^{2020}-c^{2018}=c^{2018}(c^2-1)=c^{2017}\cdot c(c-1)(c+1)\vdots6,$

=nên $c^{2020}\equiv c^{2018}\pmod6.$

Cộng ba đồng dư trên,

$a^{2018}+b^{2019}+c^{2020}\equiva^{2016}+b^{2017}+c^{2018}\pmod6.$

Theo giả thiết, $a^{2016}+b^{2017}+c^{2018}\equiv0\pmod6.$

Suy ra $a^{2018}+b^{2019}+c^{2020}\equiv0\pmod6.$

Vậy $a^{2018}+b^{2019}+c^{2020}$ chia hết cho $6.$

2.

$\displaystyleM=\frac{a^2+4a+1}{a^2+a}+\frac{b^2+4b+1}{b^2+b}+\frac{c^2+4c+1}{c^2+c}.$

Ta có $\displaystyle\frac{x^2+4x+1}{x^2+x}=\frac{x^2+x+3x+1}{x(x+1)}=1+\frac{3x+1}{x(x+1)}.$

Lại có $\displaystyle\frac{3x+1}{x(x+1)}=\frac1x+\frac2{x+1}.`$

Do đó $\displaystyle\frac{x^2+4x+1}{x^2+x}=1+\frac1x+\frac2{x+1}.$

Suy ra $\displaystyleM=3+\left(\frac1a+\frac1b+\frac1c\right)+2\left(\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\right).$

Theo bất đẳng thức Cauchy,

$\displaystyle(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge(1+1+1)^2=9.$

Vì $a+b+c\le3,$ nên $\displaystyle\frac1a+\frac1b+\frac1c\ge\frac9{a+b+c}\ge3.$

Mặt khác, áp dụng bất đẳng thức Cauchy,

$\displaystyle(a+b+c+3)\left(\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\right)\ge9.$

Do $a+b+c+3\le6,$ suy ra $\displaystyle\frac1{a+1}+\frac1{b+1}+\frac1{c+1}\ge\frac96=\frac32.$

Vậy $\displaystyleM\ge3+3+2\cdot\frac32=9.$

Dấu ``='' xảy ra khi $a=b=c=1,$ vì khi đó $a+b+c=3$ và các bất đẳng thức Cauchy đều đạt dấu bằng.

Vậy $M_{\min}=9.$

Bài 1:

Đặt 2018=a

\(B=\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}+\dfrac{a}{a+1}\)

\(=1+a-\dfrac{a}{a+1}+\dfrac{a}{a+1}=1+a=2019\)

Ta có: \(a^{2017}+b^{2017}\)= \(2a^{^{ }1018}.b^{1018}\)

⇔ (a²⁰¹⁷ + b²⁰¹⁷)² = 4(ab)²⁰¹⁸

Lại có: (a²⁰¹⁷ + b²⁰¹⁷)² ≥ 4a²⁰¹⁷.b²⁰¹⁷

⇒ 4(ab)²⁰¹⁶ ≥ 4a²⁰¹⁷.b²⁰¹⁷

⇒ ab²⁰¹⁶ ≥ ab²⁰¹⁷

⇒ ab ≤ 1

⇒ 1 - ab ≥ 0

⇒ 2018 - 2018ab ≥ 0

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