\(\Rightarrow a-1\inư\left(7\right)=\left\{1;-1;7;-7\right\}\)

\(\left[{}\begin{matrix}a-1=1\\a-1=-1\\a-1=7\\a-1=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\a=0\\a=8\\a=-6\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x=5\\y=0\end{matrix}\right.\end{matrix}\right.\)

theo bđt cauchy schwarz ta có

\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)

theo bđt cauchy schwarz ta có

\(12\sqrt{6-x}\le\dfrac{12^2+6-x}{2}\)

\(5\sqrt{x-5}\le\dfrac{25+x-5}{2}\)

\(\Rightarrow12\sqrt{6-x}+5\sqrt{x-5}\le\dfrac{144+6-x+25+x-5}{2}=85\)

vậy Amax=85

ê! Do kyung soo! teencode nhiều vậy!

S=\(2^2+4^2+....+20^2\)

\(S=2^2.1^2+2^2.2^2+2^2.3^2+.....+2^2.10^2\)

S=\(2^2\left(1^2+2^2+...+10^2\right)\)

S=4.385=1540

hay đến muốn khóc luôn nè