Đặt \(\dfrac{y+z+3}{x}=\dfrac{x+z+4}{y}=\dfrac{x+y-7}{z}=\dfrac{1}{x+y+z}=k\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+3=kx\\x+z+4=ky\\x+y-7=kz\\x+y+z=\dfrac{1}{k}\left(1\right)\end{matrix}\right.\)
Từ (1) => kx + ky + kz = 1
Suy ra y + z + 3 + x + z + 4 + x + y - 7 = 1
=> 2(x + y + z) = 1 \(\Rightarrow2.\dfrac{1}{k}=1\Rightarrow k=2\)
\(\Rightarrow\left\{{}\begin{matrix}y+z+3=2x\\x+z+4=2y\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x-y-z=3\\x-2y+z=-4\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{6}\\y=\dfrac{3}{2}\\z=-\dfrac{13}{6}\end{matrix}\right.\)
Ta có \(A=\left(6x-6\right)^{2023}+\left(2y-2\right)^{2024}+\left(6z+12\right)^{2022}\)
\(=\left(6\cdot\dfrac{7}{6}-6\right)^{2023}+\left(2\cdot\dfrac{3}{2}-2\right)^{2024}+\left(6\cdot-\dfrac{13}{6}+12\right)^{2022}\)
\(=1^{2023}+1^{2024}+1^{2022}=1+1+1=3\)
Vậy A=3
