c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)

\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)

\(=\left(2x-y+2\right)^2\)

Bài 1:

a: \(A=-x^2-4y^2+6x-8y+3\)

\(=-x^2+6x-9-4y^2-8y-4+16\)

\(=-\left(x-3\right)^2-4\left(y+1\right)^2+16\le16\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y+1=0

=>x=3 và y=-1

b: \(B=x^4-6x^3+15x^2-20x-15\)

\(=x^4-4x^3+4x^2-\left(2x^3-8x^2+8x\right)+\left(3x^2-12x+12\right)-27\)

\(=\left(x^2-4x+4\right)\left(x^2-2x+3\right)-27=\left(x-2\right)^2\cdot\left(x^2-2x+3\right)-27\ge-27\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

a: \(\dfrac{2x^3-5x^2-x+1}{2x+1}\)

\(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}\)

\(=x^2-3x+1\)

b: \(\dfrac{x^3-2x+4}{x+2}\)

\(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}\)

\(=x^2-2x+2\)

\(-B=\left(x^2-3x\right)\left(x^2-3x+10\right)-2010=\left(x^2-3x+5\right)^2-2035\).

Ta có \(x^2-3x+5=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}>0\forall x\).

Do đó \(-B\ge\left(\dfrac{11}{4}\right)^2-2035=\dfrac{-32439}{16}\Rightarrow B\le\dfrac{32439}{16}\).

...

Chọn B

\(a,n^3-2n^2+3n+3=n^3-n^2-n^2+n+2n-2+5\\ =\left(n-1\right)\left(n^2-n+2\right)+5\\ \Leftrightarrow n^3-2n^2+3n+3⋮\left(n-1\right)\\ \Leftrightarrow5⋮n-1\\ \Leftrightarrow n-1\in\left\{-5;-1;1;5\right\}\\ \Leftrightarrow n\in\left\{-4;0;2;6\right\}\)

\(b,\Leftrightarrow x^4+6x^3+7x^2-6x+a\\ =x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1-1+a\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)-1+a\\ =\left(x^2+3x-1\right)^2+a-1\)

Để \(x^4+6x^3+7x^2-6x+a⋮x^2+3x-1\)

\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)

\(B=\dfrac{2010}{4x+20\sqrt{x}+30}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}\right)^2+2\cdot2\sqrt{x}\cdot5+25+5}\)

\(B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\)

Ta có: \(\left(2\sqrt{x}+5\right)^2+5\ge5\)

\(\Rightarrow B=\dfrac{2010}{\left(2\sqrt{x}+5\right)^2+5}\le\dfrac{2010}{5}=402\)

Vậy: \(B_{min}=402\)

a) ( x 2  – 4x + 1)( x 2  – 2x + 3).

b) ( x 2  + 5x – 1)( x 2  + x – 1).