A=$\sqrt[]{-x^2+2x+4}$ = $\sqrt[]{-x^2+2x-1+5}$=$\sqrt[]{-(x-1)^2+5}$
vì -(x-1)$^{2}$ $\leq$ 0 với mọi x
<-> -(x-1)$^{2}$+5 $\leq$ 5 với mọi x
<-> A $\leq$ $\sqrt[]{5}$
<-> A max=$\sqrt[]{5}$ dấu bằng xảy ra <-> (x-1)$^{2}$=0 <-> x=1
B=$\frac{\sqrt[]{x-2}}{x}$+$\frac{\sqrt[]{y-1}}{y}$
xét $\frac{x}{\sqrt[]{x-2}}$ = $\frac{x-2+2}{\sqrt[]{x-2}}$ = $\sqrt[]{x-2}$ +$\frac{2}{\sqrt[]{x-2}}$ $\geq$ 2$\sqrt[]{2}$
<-> $\frac{\sqrt[]{x-2}}{x}$$\leq$ $\frac{\sqrt[]{2}}{4}$
xét $\frac{y}{\sqrt[]{y-1}}$ = $\frac{y-1+1}{\sqrt[]{y-1}}$ = $\sqrt[]{y-1}$ +$\frac{1}{\sqrt[]{y-1}}$ $\geq$ 2
<-> $\frac{\sqrt[]{y-1}}{y}$$\leq$ $\frac{1}{2}$
=> B$\leq$ $\frac{2+\sqrt[]{2}}{4}$
-> B max=$\frac{2+\sqrt[]{2}}{4}$ dấu bằng xảy ra <-> $\left \{ {{x-2=2} \atop {y-1=1}} \right.$ <-> $\left \{ {{x=4} \atop {y=2}} \right.$
C=$\sqrt[]{x-1}$ + $\sqrt[]{y-2}$
áp dụng bdt Bunhiacopski
(1.$\sqrt[]{x-1}$ + 1.$\sqrt[]{y-2}$ )$^{2}$ $\leq$ (1$^{2}$ +1$^{2}$ )(x-1+y-2)=2.(4-3)=2
-> C$^{2}$ $\leq$ 2 <-> C$\leq$ $\sqrt[]{2}$
Cmax=$\leq$ $\sqrt[]{2}$ dấu bằng xảy ra <->$\left \{ {{ \sqrt[]{x-1} =\sqrt[]{y-2} } \atop {x+y=4}} \right.$<->$\left \{ {{x-1=y-2} \atop {x+y=4}} \right.$ <-> $\left \{ {{x=\frac{3}{2} } \atop {y=\frac{5}{2}}} \right.$
