Toán Lớp 8: Tính GTNN: B= x^2 – 3x – 4
C= 9x^2 -42x +21
D = x^2 + 5x – 4
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Toán Lớp 8: Tính GTNN: B= x^2 – 3x – 4
C= 9x^2 -42x +21
D = x^2 + 5x – 4
Comments ( 2 )
=B=x^{2}-2. \dfrac{3}{2} x +\dfrac{9}{4} -\dfrac{3}{2} -4$
$=(x-\dfrac{3}{2})²-\dfrac{25}{4}$
Ta có:$(x-\dfrac{3}{2})²\geq0 Vói mọi x$
=>$ (x-\dfrac{3}{2})²-\dfrac{25}{4} $$\geq$ $-\dfrac{25}{4}$
Vậy MinB=$-\dfrac{25}{4}$
<=> x=$\dfrac{3}{2}$
—————
c,
$9x^{2}-42+21
=$ $9x^{2}-2.3.7.x+49-28$
= $(3x-7)^{2}-28$
Ta có :$(3x-7)^{2}\geq0$
=> $(3x-7)^{2}-28$ $\geq-28$
Vậy $Min C= -28
<=> \dfrac{7}{3}$
—————
d,
D= $x^{2}+5x-4
=$ $x^{2}+2. \dfrac{5}{2} x +\dfrac{25}{4} -\dfrac{25}{4} -4$
= $x^{2}+2. \dfrac{5}{2} x +\dfrac{25}{4} -\dfrac{41}{4}$
$=(x+\dfrac{5}{2})²-\dfrac{41}{4}$
Ta có$:(x+\dfrac{5}{2})²\geq0 với moi x$
=>$(x+\dfrac{5}{2})²-\dfrac{41}{4}\geq -\dfrac{41}{4} $
Vậy $MIn D= -\dfrac{41}{4}
<=> x=$ $-\dfrac{5}{2}$