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

Toán Lớp 9: A = $-2x^{2}$ – 3x + 5 Tìm max A

Toán Lớp 9: A = $-2x^{2}$ – 3x + 5
Tìm max A

Comments ( 2 )

  1. $A=-2x^{2}-3x+5$
    $=-2x^{2}-3x-\dfrac{9}{8}+\dfrac{49}{8}$
    $=-2(x^{2}+\dfrac{3}{2}x+\dfrac{9}{16})+\dfrac{49}{8}$
    $=-2(x+\dfrac{3}{4})^{2}+\dfrac{49}{8}$
    Ta có: $(x+\dfrac{3}{4})^{2}$ $\geq0$ 
    -> $-2(x+\dfrac{3}{4})^{2}$ $\leq0$ 
    -> $-2(x+\dfrac{3}{4})^{2}+\dfrac{49}{8}$ $\leq\dfrac{49}{8}$ 
    Dấu bằng xảy ra khi: $x^{}+\dfrac{3}{4}=0$ 
                                 ⇔ $x^{}=\dfrac{-3}{4}$ 
    Chúc bạn học tốt !!!!

  2. A = -2x^2 -3x + 5
    = -2x^2 – 3x – 9/8 + 49/8
    = -2(x^2 + 3/2x + 9/16) + 49/8
    = -2 (x^2 + 3/4)^2 + 49/8
    – Có: (x^2 + 3/4)^2 >= 0
    – Vậy: -2(x^2 + 3/4)^2 <= 0
    => -2 (x^2 + 3/4)^2 + 49/8 <= 49/8
    – Vậy bằng nhau khi: x + 3/4 = 0 
                                        x = -3/4
    @ Rin
     

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