2x2-6x=?
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\(Sửa:\left(2x^4-7x^3-7x^2-6x-2\right):\left(2x^2+x-1\right)\\ =\left(2x^4+x^3-x^2-8x^3-4x^2+4x-2x^2-x+1-9x-3\right):\left(2x^2+x-1\right)\\ =\left[x^2\left(2x^2+x-1\right)-4x\left(2x^2+x-1\right)-\left(2x^2+x-1\right)-9x-3\right]:\left(2x^2+x-1\right)\\ =x^2-4x-1\left(\text{dư }-9x-3\right)\)
\(A=-3x^2+6x-7=-3\left(x^2-2x+1-1\right)-7\)
\(=-3\left(x-1\right)^2-4\le-4\)Dấu ''='' xảy ra khi x = 1
\(B=-2x^2+5x+1=-2\left(x^2-\dfrac{5}{2}x\right)+1\)
\(=-2\left(x^2-2.\dfrac{5}{4}x+\dfrac{25}{16}-\dfrac{25}{16}\right)+1\)
\(=-2\left(x-\dfrac{5}{4}\right)^2+\dfrac{33}{8}\le\dfrac{33}{8}\)Dấu ''='' xảy ra khi x = 5/4
C;D chỉ có GTNN thôi bạn nhé \(C=2x^2-8x+13=2\left(x^2-4x+4-4\right)+13\)
\(=2\left(x-2\right)^2+5\ge5\)Dấu ''='' xảy ra khi x = 2
\(D=x^2-3x+5=x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}-\dfrac{9}{4}+5\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)Dấu ''='' xảy ra khi x = 3/2
d: Ta có: \(D=x^2-3x+5\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{11}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
-6x(2x2 - x + \(\dfrac{2}{3}\))
= (-6x) . 2x2 - (-6x) . x + (-6x) . \(\dfrac{2}{3}\)
= -12x3 - (-6x2) + (-4x)
= -12x3 + 6x2 - 4x
\(2x^2-6x+8=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}+8=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{2}\)
Vì \(2\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{2}\)
Vậy GTNN của \(2x^2-6x+8\) là \(\dfrac{7}{2}\) khi và chỉ khi \(x=\dfrac{3}{2}\)
\(\left(x^2+6x-1\right)^2+2x^2+x^4+2\left(x^2+6x-1\right)\left(x^2+1\right)\)
\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+\left(x^2+1\right)^2-1=\left(x^2+6x-1+x^2+1\right)^2-1=\left(2x^2+6x\right)^2-1=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)
\(\left(x^2+6x-1\right)^2+2\left(x^2+6x-1\right)\left(x^2+1\right)+x^4+2x^2\)
\(=\left(x^2+6x-1\right)\left(x^2+6x-1+2x^2+2\right)+x^4+2x^2\)
\(=\left(x^2+6x-1\right)\left(3x^2+6x+1\right)+x^4+2x^2\)
\(=\left(2x^2+6x-1\right)\left(2x^2+6x+1\right)\)
`2x^2+6x=2(x^2+3x)=2(x^2+2 . x . 3/2 + 9/4)-9/2=2(x+3/2)^2-9/2`
Ta thấy : `2(x+3/2)^2>=0`
`->2(x+3/2)^2-9/2>=-9/2`
Dấu = xảy ra `<=>x+3/2=0` `<=>x=-3/2`
vậy GTNN của biểu thức là `-9/2` khi `x=-3/2`
Ta có: 2x2-6x=\(2\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" xảy ra <=> \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
\(N=-\dfrac{1}{2}x^2+6x-20=-\dfrac{1}{2}\left(x^2-12x+36\right)-2\)
\(=-\dfrac{1}{2}\left(x-6\right)^2-2\le-2\)
\(maxN=-2\Leftrightarrow x=6\)
\(2x^2-6x=\left(2x\right)\cdot x-\left(2x\right)\cdot3=2x\left(x-3\right)\)
\(2x^2-6x=2x\left(x-3\right)\)