Xét Max
Áp dụng bất đẳng thức Bunhiacopxki
`(1.\sqrt{x-4}+1.\sqrt{y-3})^2≤(1+1)(x-4+y-3)=2.8=16`
`⇒\sqrt{x-4}+\sqrt{y-3}≤4`
`⇒B≤4`
Dấu `=` xảy ra khi `\sqrt{x-4}=\sqrt{y-3} `
`⇔x-4=y-3`
`⇔x+y=y+y-3+4`
`⇔15=2y+1`
`⇔14=2y`
`⇔y=7`
`⇒x=8`
Xét Min
`(\sqrt{x-4}+\sqrt{y-3})^2=x+y-4-3+2\sqrt{(x-4)(y-3)}=8+2\sqrt{(x-4)(y-3)}`
Vì `2\sqrt{(x-4)(y-3)}≥0`
`⇒B^2≥8+0=8`
`⇒B≥2\sqrt{2}`
Dấu `=` xảy ra khi `(x-4)(y-3)=0`
`⇔`\(\left[ \begin{array}{l}\left \{ {{x-4=0} \atop {15-x=y}} \right.\\\left \{ {{y-3=0} \atop {15-y=x}} \right.\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}\left \{ {{x=4} \atop {y=11}} \right.\\\left \{ {{y=3} \atop {x=12}} \right.\end{array} \right.\)
