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

Toán Lớp 8: tìm GTLN: N= 2x ²+5x-3y ²-y+20

Toán Lớp 8: tìm GTLN:
N= 2x ²+5x-3y ²-y+20

Comments ( 2 )

  1. Lời giải và giải thích chi tiết:
    N=- 2x^2+5x-3y^2-y+20
    =(-2x^2+5x-25/8)+(-3y^2-y-1/12)+557/24
    =-2(x^2-5/2x+25/16)-3(y^2+1/3y+1/36)+557/24
    =-2(x-5/4)^2-3(y+1/6)^2+557/224<=557/224AAx,y
    Dấu = xảy ra <=>$\left\{\begin{matrix}x-\dfrac54=0\\y+\dfrac16=0\end{matrix}\right.$=>$\left\{\begin{matrix}x=\dfrac54\\y=-\dfrac16\end{matrix}\right.$
          Vậy Nmax=557/224<=>$\left\{\begin{matrix}x=\dfrac54\\y=-\dfrac16\end{matrix}\right.$

  2. N=-2x^2 + 5x – 3y^2 – y + 20
    =-2(x^2 – 2. x . 5/4 + 25/16)- 3(y^2 + 2. y . 1/6 + 1/36) + 557/24
    =>-2(x-5/4)^2 – 3(y+1/6)^2 + 407/24 >= 557/24
    => Dấu “=” xảy ra khi:
    $\begin{cases} x- \dfrac{5}{4}=0 \\y+\dfrac{1}{6}=0 \end{cases}$ => $\begin{cases} x=\dfrac{5}{4}\\y=-\dfrac{1}{6} \end{cases}$ 
    Vậy N max = 557/24 <=> $\begin{cases} x=\dfrac{5}{4}\\y=-\dfrac{1}{6} \end{cases}$ 

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