Đáp án :
`A_(min)=22` khi `x=1; y=2`
Giải thích các bước giải :
`A=x^2+2y^2+1/x+24/y`
`=>A=(x^2-2x+1)+(2y^2-8y+8)+(x-2+1/x)+(6y-24+24/y)+(x+2y)+17`
`=>A=(x^2-2x+1)+2.(y^2-4y+4)+(x^2/x-(2x)/x+1/x)+6.(y^2/y-(4y)/y+4/y)+(x+2y)+17`
`=>A=(x^2-2x+1)+2.(y^2-4y+4)+(x^2-2x+1)/x+(6.(y^2-4y+4))/y+(x+2y)+17`
`=>A=(x-1)^2+2.(y-2)^2+(x-1)^2/x+(6.(y-2)^2)/y+(x+2y)+17`
Vì `(x-1)^2>=0;2.(y-2)^2>=0;(x-1)^2/x>=0;(6.(y-2)^2)/y>=0;x+2y>=5`
`=>(x-1)^2+2.(y-2)^2+(x-1)^2/x+(6.(y-2)^2)/y+(x+2y)+17>=17+5=22`
`=>A>=22`
`=>A_(min)=22`
Xảy ra dấu `=` khi : `x=1; y=2`
Vậy : `A_(min)=22` khi `x=1; y=2`
