Toán Lớp 8: Tìm MinA=x^2-4x+y^2+6y+15 Tìm MaxB=(3-x)(x+5)

Question

Toán Lớp 8:  Tìm MinA=x^2-4x+y^2+6y+15 Tìm MaxB=(3-x)(x+5)

haiyemy · Answer

a) A = x^2 - 4x + y^2 + 6y + 15 = (x^2 - 4x + 4) + (y^2 + 6y +9) + 2 = (x^2 - 2 . x . 2 + 2^2) + (y^2 + 2 . y . 3 + 3^2) + 2 = (x-2)^2 + (y+3)^2 + 2 forall x ; y ta có : (x-2)^2 ge 0 (y+3)^2 ge 0 => (x-2)^2+ (y+3)^2 ge 0 => (x-2)^2 + (y+3)^2 + 2 ge 2 => A ge 2 Dấu = xảy ra <=> {(x-2=0),(y+3=0):} <=> {(x =2),(y=-3):} Vậy text{Min}_A = 2 <=> {(x =2),(y=-3):} b) B = (3-x)(x+5) = 3x + 15 - x^2 - 5x = -x^2 - 2x + 15 = -(x^2 + 2x + 1) + 16 = -(x^2 + 2 . x . 1 + 1^2) + 16 = -(x+1)^2 + 16 forall x ta có : (x+1)^2 ge 0 => -(x+1)^2 le 0 => -(x+1)^2+ 16 le 16 => B le 16 Dấu = xảy ra <=> x+1=0<=>x=-1 Vậy text{Max}_B = 16 <=> x = -1

thanoanghha · Answer

A=x^2-4x+y^2+6y+15   A=x^2-4x+4+y^2+6y+9+2   A=(x^2-4x+4)+(y^2+6y+9)+2   A=(x-2)^2+(y+3)^2+2   Ta c&oacute;:   (x-2)^2&ge;0&forall;x   (y+3)^2&ge;0&forall;y   &rArr;(x-2)^2+(y+3)^2+2&ge;2&forall;x,y   Vậy min A bằng 2 khi x=2;y=-3   $  $   B=(3-x)(x+5)   B=3x+15-x^2-5x   B=-x^2-2x+15   B=-(x^2-2x+15)   B=-(x^2-2x+1)+16   B=-(x-1)^2+16   Ta c&oacute;:   (x-1)^2&ge;0&forall;x   &rArr;-(x-1)^2&le;0&forall;x   &rArr;-(x-1)^2+16&le;16&forall;x   Vậy max B bằng 16 khi x-1=0&hArr;x=1