Lời giải:

ĐKĐB \(\Leftrightarrow a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}\)

\(\Rightarrow \left\{\begin{matrix} a-b=\frac{b-c}{bc}\\ b-c=\frac{c-a}{ac}\\ c-a=\frac{a-b}{ab}\end{matrix}\right.\)

\(\Rightarrow (a-b)(b-c)(c-a)=\frac{(b-c)(c-a)(a-b)}{a^2b^2c^2}\)

Vì $a,b,c$ đôi 1 khác nhau nên $a^2b^2c^2=1$. Khi đó:

\(P=(5.1^3-8.1+2)^{2020}=(-1)^{2020}=1\)

Đặt \(\dfrac{a}{b^2}=\dfrac{b^2}{c^3}=\dfrac{c^3}{a^4}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=k.b^2\\b^2=k.c^3\\c^3=k.a^4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=k.k.c^3=k^2c^3\\c^3=k.a^4\end{matrix}\right.\)

\(\Rightarrow a=k^2.k.a^4\)

\(\Rightarrow a=k^3a^4\)

\(\Rightarrow\left(ka\right)^3=1\)

\(\Rightarrow ka=1\)

\(\Rightarrow a=\dfrac{1}{k}\) (1)

Thế vào \(c^3=k.a^4\Rightarrow c^3=k.\dfrac{1}{k^4}=\dfrac{1}{k^3}\)

\(\Rightarrow c=\dfrac{1}{k}\) (2)

Thế vào \(b^2=kc^3\Rightarrow b^2=k.\dfrac{1}{k^3}=\dfrac{1}{k^2}\)

\(\Rightarrow b=\dfrac{1}{k}\) hoặc \(b=-\dfrac{1}{k}\) (3)

(1);(2);(3) \(\Rightarrow\left[{}\begin{matrix}a=b=c\\a=c=-b\end{matrix}\right.\)

TH1: \(a=b=c\)

\(\Rightarrow P=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)

Th2: \(a=c=-b\)

\(\Rightarrow P=\left(1+\dfrac{-b}{b}\right)\left(1+\dfrac{b}{-b}\right)\left(1+\dfrac{-b}{-b}\right)=0.0.2=0\)

Chọn B. Thay \(\dfrac{1}{3}\)vào x và \(\dfrac{1}{2}\)vào y 

giải để ra được m

\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)

Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

b) Ta có: \(4x^2+x-5=0\)

\(\Leftrightarrow4x^2-4x+5x-5=0\)

\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)

Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:

\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)

Vậy: Khi \(4x^2+x-5=0\) thì B=0

Từ x=\(\dfrac{1}{2}\)a+\(\dfrac{1}{2}\)b+\(\dfrac{1}{2}\)c=\(\dfrac{1}{2}\).(a+b+c)\(\Rightarrow\)2x=(a+b+c)

M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x\(^2\)

= x\(^2\)-xb-ax+ab+x\(^2\)-xc-bx+bc+x\(^2\)-ax-cx+ac+x\(^2\)

= 4x\(^2\)-2ac-2bx-2cx+ab+bc+ac

= 4x\(^2\)-2x(a+b+c)+ab+bc+ca

Thay 2x=a+b+c,ta được:

M= 4x\(^2\)-2x.2c+ab+bc+ca

M= 4x\(^2\)-4x\(^2\)+ab+bc+ca

M= ab+bc+ca