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\(a,n^3-2n^2+3n+3=n^3-n^2-n^2+n+2n-2+5\\ =\left(n-1\right)\left(n^2-n+2\right)+5\\ \Leftrightarrow n^3-2n^2+3n+3⋮\left(n-1\right)\\ \Leftrightarrow5⋮n-1\\ \Leftrightarrow n-1\in\left\{-5;-1;1;5\right\}\\ \Leftrightarrow n\in\left\{-4;0;2;6\right\}\)
\(b,\Leftrightarrow x^4+6x^3+7x^2-6x+a\\ =x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1-1+a\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)-1+a\\ =\left(x^2+3x-1\right)^2+a-1\)
Để \(x^4+6x^3+7x^2-6x+a⋮x^2+3x-1\)
\(\Leftrightarrow a-1=0\Leftrightarrow a=1\)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
\(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2-6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2+3x-1\right)^2\)
Lời giải:
$A=x^4-4x^3+7x^2-12x+75$
$=(x^2-2x)^2+3x^2-12x+75$
$=(x^2-2x)^2+3(x^2-4x+4)+63$
$=(x^2-2x)^2+3(x-2)^2+63\geq 63$
Vậy $A_{\min}=63$. Giá trị này đạt tại $x^2-2x=x-2=0$
$\Leftrightarrow x=2$
\(A=\left(x^4-4x^3+4x^2\right)+\left(3x^2-12x+12\right)+63\)
\(A=x^2\left(x^2-4x+4\right)+3\left(x^2-4x+4\right)+63\)
\(A=\left(x^2+3\right)\left(x-2\right)^2+63\ge63\)
\(A_{min}=63\) khi \(x=2\)
\(3,=\left(x-y\right)^3+\left(y-x+x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3+\left(y-x\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-x+x-z\right)+\left(x-z\right)^3+\left(z-x\right)^3\\ =\left(x-y\right)^3-\left(x-y\right)^3+3\left(y-x\right)\left(x-z\right)\left(y-z\right)-\left(z-x\right)^3+\left(z-x\right)^3\\ =3\left(y-x\right)\left(x-z\right)\left(y-z\right)\)
\(4,=\left(x^4+3x^3-x^2\right)+\left(3x^3+9x^2-3x\right)-\left(x^2+3x-1\right)\\ =x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)\left(x^2+3x-1\right)\\ =\left(x^2+3x-1\right)^2\)
