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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: Tìm GTNN của: `S = x^2 + xy + y^2 – 3x – 3y + 15`

Toán Lớp 8: Tìm GTNN của: S = x^2 + xy + y^2 – 3x – 3y + 15

Comments ( 2 )

  1. A=x^2+xy+y^2-3x-3y+15
    =(x^2+xy-3x)+y^2-3y+15
    =[x^2+2x(1/2y-3/2)+(1/2y-3/2)^2]-(1/2y-3/2)^2+y^2-3y+15
    =(x+1/2y-3/2)^2-1/4y^2+3/2y-9/4+y^2-3y+15
    =(x+1/2y-3/2)^2+3/4y^2-3/2y+51/4
    =(x+1/2y-3/2)^2+3/4y^2-3/2y+3/4+12
    =(x+1/2y-3/2)^2+3/4(y^2-2y+1)+12
    =(x+1/2y-3/2)^2+3/4(y-1)^2+12\ge 12
    Đẳng thức xảy ra <=>x=y=1 
    Vậy min A=12 đạt được khi x=y=1

  2. $S=x^{2}+xy+y^{2}-3x-3y+15$
    $S=x^{2}+xy-3x+y^{2}-3y+15$
    $S=x^{2}+2x(\frac{1}{2}y-\frac{3}{2})+(\frac{1}{2}y-\frac{3}{2})^{2}-(\frac{1}{2}y-\frac{3}{2})^{2}+y^{2}-3y+15$
    $S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}-\frac{1}{4}y^{2}+\frac{3}{2}y-\frac{9}{4}-3y+15$
    $S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}y^{2}-\frac{3}{2}y+\frac{51}{4}$
    $S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}y^{2}-\frac{3}{2}y+\frac{3}{4}+12$
    $S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}(y^{2}-2y+1)+12$
    $S=[x+(\frac{1}{2}y+\frac{3}{2})]^{2}+\frac{3}{4}(y-1)^{2}+12\geq12$
    Dấu $”=”$ xảy ra
    <=>$\left \{ {{x+(\frac{1}{2}y+\frac{3}{2})=0} \atop {y-1=0}} \right.$ 
    <=>$\left \{ {{x+(\frac{1}{2}.1+\frac{3}{2})=0} \atop {y=1}} \right.$ 
    <=>$\left \{ {{x=1} \atop {y=1}} \right.$ 
    <=>$x=y=1$
    Vậy $MinS=12$ khi $x=y=1$

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222-9+11+12:2*14+14 = ? ( )

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