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

Toán Lớp 8: tìm min(max) A = 2x^2 + 5y^2 – 4xy – 26y + 20x + 60

Toán Lớp 8: tìm min(max)
A = 2x^2 + 5y^2 – 4xy – 26y + 20x + 60

Comments ( 1 )

  1. ~ gửi bạn ~
    Giải đáp
    $min_A=7 \Leftrightarrow \left\{\begin{array}{l} x=-4\\y=1\end{array} \right..$
    Lời giải và giải thích chi tiết:
    $A=2x^2+5y^2-4xy-26y+20x+60\\ =2\left(x^2+\dfrac{5}{2}y^2-2xy-13y+10x+30\right)\\ =2\left(x^2-2xy+y^2+\dfrac{3}{2}y^2-13y+10x+30\right)\\ =2\left[(x-y)^2+10x-10y+25+\dfrac{3}{2}y^2-3y+\dfrac{3}{2}+\dfrac{7}{2}\right]\ =2\left[(x-y)^2+10(x-y)+25+\dfrac{3}{2}\left(y^2-2y+1\right)+\dfrac{7}{2}\right]\\\ =2\left[(x-y+5)^2+\dfrac{3}{2}\left(y^2-2y+1\right)+\dfrac{7}{2}\right]\\ =2(x-y+5)^2+3\left(y-1\right)^2+7 \ge 7 \ \forall \ x$ Dấu “=” xảy ra $\Leftrightarrow \left\{\begin{array}{l} x-y+5=0\\y-1=0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x=-4\\y=1\end{array} \right.$
    Vậy $min_A=7 \Leftrightarrow \left\{\begin{array}{l} x=-4\\y=1\end{array} \right..$  

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

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