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

Toán Lớp 8: tìm giá trị lớn nhất `(1 – 3x)(x + 2)`

Toán Lớp 8: tìm giá trị lớn nhất (1 – 3x)(x + 2)

Comments ( 2 )

  1. Ta có :
    (1-3x)(x+2)
    =  -3x^2 -5x+2
    = -3(x^2 +5/3x -2/3)
    = -3(x^2 + 2 . x . 5/6 + 25/36    – 49/36)
    = -3(x+5/6)^2 +49/12 \le 49/12 AA x 
    $\\$
    Vậy Max = 49/12 <=> x+5/6 = 0 <=> x = -5/6.

  2. Gửi bạn:
    $(1-3x)(x+2)$
    $=x+2-3x^2-6x$
    $=-3x^2-5x+2$
    $=-3(x^2+\dfrac{5x}{3}-\dfrac{2}{3})$
    $=-3(x^2+\dfrac{5x}{3}+\dfrac{25}{36}-\dfrac{49}{36})$
    $=-3(x+\dfrac{5}{6})^2+3.\dfrac{49}{36}$
    $=-3(x+\dfrac{5}{6})^2+\dfrac{49}{12}$
    Ta có: $-3(x+\dfrac{5}{6})^2+\dfrac{49}{12}≤\dfrac{49}{12}$
    $⇒$ $Max_A=\dfrac{49}{12}$
    Dấu $’=’$ xảy ra khi:
    $-3(x+\dfrac{5}{6})^2=0$
    $⇒x+\dfrac{5}{6}=0$
    $⇒x=-\dfrac{5}{6}$
    Vậy $Max_A=\dfrac{49}{12}$ khi $x=-\dfrac{5}{6}$

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

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