Xét phương trình có dạng $ax^2+bx+c=0$ có: \(\left\{{}\begin{matrix}a=1\ne0\\b=-\left(m+1\right)\\c=m\end{matrix}\right.\)

suy ra phương trình là phương trình bậc 2 một ẩn x
Có \(\Delta=b^2-4ac=m^2+2m+1-4.1.m=m^2-2m+1=\left(m-1\right)^2\ge0\)

nên phương trình luôn có 2 nghiệm 
Theo hệ thức Vi-et \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left[-\left(m+1\right)\right]}{1}=m+1\\x_1.x_2=\dfrac{c}{a}=\dfrac{m}{1}=m\end{matrix}\right.\)

Phương trình có 2 nghiệm trái dấu và nghiệm dương > trị tuyệt đối nghiệm âm \(\Leftrightarrow\left\{{}\begin{matrix}\Delta\ge0\\ac< 0\\x_1+x_2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m+1>0\end{matrix}\right.\Leftrightarrow0>m>-1\)

\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)

\(\dfrac{1}{13}-\dfrac{1}{18}=\dfrac{18}{13.18}-\dfrac{13}{13.18}=\dfrac{18-13}{13.18}=\dfrac{5}{13.18}\)

\(\dfrac{1}{n}-\dfrac{1}{n+k}=\dfrac{n+k}{n\left(n+k\right)}-\dfrac{n}{n\left(n+k\right)}=\dfrac{n+k-n}{n\left(n+k\right)}=\dfrac{k}{n\left(n+k\right)}\)

\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)

\(\Leftrightarrow8x^9+2x^3=x^3+3x^2+3x+1+x+1\)

\(\Leftrightarrow8x^9+2x^3=\left(x+1\right)^3+\left(x+1\right)\)

Đặt $2x^3=a;x+1=b$ có:

\(a^3+a=b^3+b\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)

\(\Leftrightarrow a=b\left(doa^2+b^2+ab+1>0\right)\)

hay \(2x^3=x+1\)

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

\(\Leftrightarrow x=1\left(2x^2+2x+1=x^2+\left(x+1\right)^2>0\forall x\right)\)

Vậy $x=1$ t/m đề
 

Có $\dfrac{a^2}{5a^2+(b+c)^2}=\dfrac{1}{9}.\dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$

Áp dụng Cauchy-Schwarz có:

$\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{4a^2}{2a(2a+bc)} \geq \dfrac{9a^2}{a^2+b^2+c^2+2a(2a+bc)}$

Nên $\dfrac{a^2}{5a^2+(b+c)^2} \leq \dfrac{1}{9}.(\dfrac{a^2}{a^2+b^2+c^2}+\dfrac{2a}{2a+bc})$

Tương tự $\dfrac{b^2}{5b^2+(a+c)^2} \leq \dfrac{1}{9}.(\dfrac{b^2}{a^2+b^2+c^2}+\dfrac{2b}{2b+ac})$

 $\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(\dfrac{c^2}{a^2+b^2+c^2}+\dfrac{2c}{2c+ab})$

Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-(\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab}))$

Áp dụng Cauchy Schwarz có:

$\dfrac{bc}{2a+bc}+\dfrac{ca}{2b+ac}+\dfrac{ab}{2c+ab} \geq \dfrac{(ab+bc+ca)^2}{(ab)^2+(bc)^2+(ca)^2+2a^2bc+2ab^2c+2abc^2}=\dfrac{(ab+bc+ca)^2}{(ab+bc+ca)^2}=1$

Nên $\dfrac{a^2}{5a^2+(b+c)^2}+\dfrac{b^2}{5b^2+(a+c)^2} +\dfrac{c^2}{5c^2+(a+b)^2} \leq \dfrac{1}{9}.(1+3-1)=\dfrac{1}{3}$

Dấu $=$ xảy ra khi $a=b=c$

\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

Phương trình trên vế $a$ sẽ là $+6$

Vế $b$ sẽ là $-3$

do phương trình có dạng $ax+b=0$

Chọn $A.$ bù nhau do $128^o+52^o=180^o$