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

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

=>\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

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

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

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

=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow a=b=c\)

\(A=\dfrac{a^{2023}}{b^{2023}}+\dfrac{b^{2023}}{c^{2023}}+\dfrac{c^{2023}}{a^{2023}}\)

\(=\dfrac{a^{2023}}{a^{2023}}+\dfrac{b^{2023}}{b^{2023}}+\dfrac{c^{2023}}{c^{2023}}\)

=1+1+1

=3

a)
\(16x^2-\left(4x-5\right)^2=15\)
\(\left(4x-4x+5\right)\left(4x+4x-5\right)=15\)
\(5\left(8x-5\right)=15\)
40x-25=15
40x=40
x=1

 

b: Ta có: \(\left(2x+3\right)^2-4\left(x-1\right)\left(x+1\right)=49\)

\(\Leftrightarrow4x^2+12x+9-4x^2+16=49\)

\(\Leftrightarrow12x=24\)

hay x=2

d: Ta có: \(2\left(x+1\right)^2-\left(x-3\right)\left(x+3\right)-\left(x-4\right)^2=0\)

\(\Leftrightarrow2x^2+4x+2-x^2+9-x^2+8x-16=0\)

\(\Leftrightarrow12x=5\)

hay \(x=\dfrac{5}{12}\)