- Ta có : \(a^3+b^3+c^3=3abc\)

=> \(a^3+b^3+c^3-3abc=0\)

=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Mà \(a+b+c\ne0\)

=> \(a^2+b^2+c^2-ab-bc-ac=0\)

=> \(\frac{\left(a^2-2ab+b^2\right)+\left(b^2-2ac+c^2\right)+\left(c^2-2ac+a^2\right)}{2}=0\)

=> \(\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{2}=0\)

=> \(a-b=b-c=c-a=0\)

=> \(a=b=c\)

- Thay a = b = c vào biểu thức N ta được :

\(N=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy giá trị của N = \(\frac{1}{3}\) khi \(a^3+b^3+c^3=3abc\) và \(a+b+c\ne0\)

a: \(A=\frac{3}{\sqrt{x}-2}-\frac{2\sqrt{x}+3}{x-4}\)

\(=\frac{3\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{3\sqrt{x}+6-2\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(P=A\cdot B\)

\(=\frac{\sqrt{x}+3}{x-4}\cdot\frac{x-4}{\sqrt{x}}=\frac{\sqrt{x}+3}{\sqrt{x}}\)

b: \(P=\frac{\sqrt{x}+7}{2}\)

=>\(\frac{\sqrt{x}+3}{\sqrt{x}}=\frac{\sqrt{x}+7}{2}\)

=>\(\sqrt{x}\left(\sqrt{x}+7\right)=2\left(\sqrt{x}+3\right)\)

=>\(x+5\sqrt{x}-6=0\)

=>\(\left(\sqrt{x}+6\right)\left(\sqrt{x}-1\right)=0\)

=>\(\sqrt{x}-1=0\)

=>\(\sqrt{x}=1\)

=>x=1(nhận)

c: \(P^2-P=P\left(P-1\right)=\frac{\sqrt{x}+3}{\sqrt{x}}\left(\frac{\sqrt{x}+3}{\sqrt{x}}-1\right)\)

\(=\frac{\sqrt{x}+3}{\sqrt{x}}\cdot\frac{3}{\sqrt{x}}>0\)

=>\(P^2>P\)

Ta có 1 + ab² \(\ge\)\(2b\sqrt{a}\)

1 + bc² \(\ge2c\sqrt{b}\)

1 + ca² \(\ge2a\sqrt{c}\)

VT \(\ge\)\(2\left(\frac{b\sqrt{a}}{c^3}+\frac{c\sqrt{b}}{a^3}+\frac{a\sqrt{c}}{b^3}\right)\)

^{\(\ge2\frac{\left(\sqrt[4]{b^2a}+\sqrt[4]{c^2b}+\sqrt[4]{a^2c}\right)^2}{a^3+b^3+c^3}\)}

^{\(\ge2\frac{\left(3\sqrt[12]{a^3b^3c^3}\right)^2}{a^3+b^3+c^3}\)}

^{\(\ge\frac{18}{a^3+b^3+c^3}\)}