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

Toán Lớp 8: Tìm giá trị lớn nhất của: F = 5 – 4x^2 + 4 ; G= -4x^2+8x-2003 HELP MEEEEEEEEEEE!!!!!!

Toán Lớp 8: Tìm giá trị lớn nhất của: F = 5 – 4x^2 + 4 ; G= -4x^2+8x-2003
HELP MEEEEEEEEEEE!!!!!!

Comments ( 2 )

  1. Giải đáp:
    \(\begin{array}{l}
    a)\quad \max F = 6 \Leftrightarrow x = \dfrac12\\
    b)\quad \max G = -1999 \Leftrightarrow x = 1
    \end{array}\)
    Lời giải và giải thích chi tiết:
    \(\begin{array}{l}
    a)\quad F =5 – 4x^2 + 4x\\
    \to F = -(4x^2 – 4x + 1) + 6\\
    \to F = – (2x – 1)^2 + 6\\
    \text{Ta có:}\\
    \quad (2x-1)^2\geqslant 0\quad \forall x\\
    \to – (2x-1)^2 \leqslant 0\\
    \to – (2x-1)^2 +6\leqslant 6\\
    \to F \leqslant 6\\
    \text{Dấu = xảy ra} \Leftrightarrow 2x – 1 =0\Leftrightarrow x = \dfrac12\\
    \text{Vậy}\ \max F = 6 \Leftrightarrow x = \dfrac12\\
    b)\quad G = -4x^2 +8x – 2003\\
    \to G = -4(x^2 – 2x + 1) – 1999\\
    \to G = -4(x-1)^2 – 1999\\
    \text{Ta có:}\\
    \quad (x-1)^2\geqslant 0\quad \forall x\\
    \to – 4(x-1)^2 \leqslant 0\\
    \to – 4(x-1)^2  -1999\leqslant -1999\\
    \to G\leqslant -1999\\
    \text{Dấu = xảy ra}\ \Leftrightarrow x – 1= 0\Leftrightarrow x = 1\\
    \text{Vậy}\ \max G = -1999 \Leftrightarrow x = 1
    \end{array}\) 

  2. $\\$
    F=5-4x^2+4
    = -4x^2 + 9\ge 9∀x
    Dấu “=” xảy ra khi :
    x^2=0↔x=0
    Vậy max F=9↔x=0
    G=-4x^2+8x-2003
    = -(4x^2 – 8x + 2003)
    =-[(2x)^2-2.2x.2 +2^2 + 1999]
    = – (2x-2)^2 – 1999\ge -1999∀x
    Dấu “=” xảy ra khi :
    (2x-2)^2=0↔x=1
    Vậy max G=-1999↔x=1
     

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

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