$a)(3x.1-2y.1)^2 \le ((3x)^2+(-2y)^2)(1^2+1^2)=2(9x^2+4^2)=2\\ =>|3x-2y| \le \sqrt{2} \forall x,y:9x^2+4y^2=1\\ b)(2x+3y)^2=(\sqrt{2}x.\sqrt{2}+\sqrt{3}y.\sqrt{3})^2 \le ((\sqrt{2}x)^2+(\sqrt{3}y)^2)((\sqrt{2})^2+(\sqrt{2})^2)=35\\ =>|2x+3y| \le \sqrt{35} \forall x,y:2x^2+3y^2=7\\ c)(4x-3y)^2=(2x.2-\sqrt{3}y.\sqrt{3})^2 \le ((2x)^2+(-\sqrt{3}y)^2)(2^2+\sqrt{3}^2)\\ <=>9\le 7(4x^2+3y^2)\\ <=>4x^2+3y^2 \ge \dfrac{9}{7}$
Dấu "=" xảy ra $<=> \dfrac{2x}{2}=\dfrac{\sqrt{3}y}{-\sqrt{3}y}<=>x=-y\\ 4x-3y=3=>x=\dfrac{3}{7};y=-\dfrac{3}{7}\\ d)y=\sqrt{x-1}+\sqrt{5-x} \le (\sqrt{x-1}^2+\sqrt{5-x}^2)(1^2+1^2)=8$
Dấu "=" xảy ra $<=> \sqrt{x-1}^2=\sqrt{5-x}<=>x=3$
