Toán Lớp 8: Tìm giá trị lớn nhất
A= -3x(x+3)-7
B= -2x^2 +5x-8
C=3-x^2+4x
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Toán Lớp 8: Tìm giá trị lớn nhất
A= -3x(x+3)-7
B= -2x^2 +5x-8
C=3-x^2+4x
Gỉai chi tiết nhé
Comments ( 2 )
$A=-3x^2-9x-7$
$A=-3x^2-9x-\dfrac{27}{4}-\dfrac{1}{4}$
$A=-3(x^2+3x+\dfrac{9}{4})-\dfrac{1}{4}$
$A=-3(x+\dfrac{3}{2})^2-\dfrac{1}{4}$
$⇔-3(x+\dfrac{3}{2})^2-\dfrac{1}{4}≤\dfrac{-1}{4}$
Dấu $”=”$ xảy ra khi $x+\dfrac{3}{2}=0⇔x=\dfrac{-3}{2}$
Vậy $A_{max}=\dfrac{-1}{4}⇔x=\dfrac{-3}{2}$
$B=-2x^2+5x-\dfrac{25}{8}-\dfrac{39}{8}$
$B=-2(x^2-\dfrac{5}{2}x+\dfrac{25}{16})-\dfrac{39}{18}$
$-2(x-\dfrac{5}{4})^2≤0∀x$
$⇔-2(x-\dfrac{5}{4})^2-\dfrac{39}{18}≤\dfrac{-39}{18}$
Dấu $”=”$ xảy ra khi $x-\dfrac{5}{4}=0⇔x=\dfrac{5}{4}$
Vậy $B_{max}=\dfrac{-39}{18}⇔x=\dfrac{5}{4}$
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$C=-x^2+4x-4+7$
$C=-(x-2)^2+7$
$-(x-2)^2≤0∀x$
$⇔-(x-2)^2+7≤7$
Dấu $”=”$ xảy ra khi $x-2=0⇔x=2$
Vậy $C_{max}=7⇔x=2$