\(A=-x^2+4x-5\\ A=-\left(x^2-4x+4\right)-1\\ A=-\left(x^2-2\cdot x\cdot2+2^2\right)-1\\ A=-\left(x-2\right)^2-1\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow-\left(x-2\right)^2\le0\forall x\\ \Leftrightarrow A=-\left(x-2\right)^2-1\le-1\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Max\right)}=-1\text{ }khi\text{ }x=2\)
\(B=-2x^2-6x+5\\ B=-2x^2-6x-\dfrac{9}{2}+\dfrac{19}{2}\\ B=-\left(2x^2+6x+\dfrac{9}{2}\right)+\dfrac{19}{2}\\ B=-2\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{2}\\ B=-2\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{19}{2}\\ B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\\ Do\text{ }\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow-2\left(x+\dfrac{3}{2}\right)^2\le0\forall x\\ \Leftrightarrow B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\le\dfrac{19}{2}\forall x\\ \text{Dấu }"="\text{ xảy }ra\text{ }khi:\\ \left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }B_{\left(Max\right)}=\dfrac{19}{2}\text{ }khi\text{ }x=-\dfrac{3}{2}\)
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Lời giải:
(A=-x^2+4x-5)
(A=-left(x^2-4x+5 ight))
(A=-left(x^2-4x+4+1 ight))
(A=-left(x^2-4x+4 ight)-1)
(A=-left(x-2 ight)^2-1le1)
Dấu "=" xảy ra khi: (x=2)
(B=-2x^2-6x+5)
(B=-2left(x^2+3x-dfrac{5}{2} ight))
(B=-2left(x^2+3x-dfrac{19}{4}+dfrac{9}{4} ight))
(B=-2left(x^2+3x+dfrac{9}{4} ight)+dfrac{19}{2})
(B=-2left(x+dfrac{3}{2} ight)^2+dfrac{19}{2}ledfrac{19}{2})
Dấu "=" xảy ra khi : (x=-dfrac{3}{2})
A = -x2 + 4x - 5
= -(x2-4x+5)
= -(x2-2x.2+4-1)
= -(x-2)2 + 1
= 1 - (x-2)2 (le) 1
=> MaxA = 1
Dấu "=" xảy ra khi x = 2
b) -2x2 - 6x + 5
= -2(x2+3x)+5
= -2(x2+2x.(dfrac{3}{2})+(dfrac{9}{4})) + (dfrac{9}{2}) + 5
= (dfrac{19}{2}) - 2(x+(dfrac{3}{2}))2 (le) (dfrac{19}{2})
=> MaxB = (dfrac{19}{2})
Dấu "=" xảy ra khi x = (-dfrac{3}{2})
