\(x^3+3x^2-4\)

\(=x^4-x^2+4x^2-4\)

\(=x^2.\left(x^2-1\right)+4.\left(x^2-1\right)\)

\(=\left(x^2+4\right).\left(x^2-1\right)\)

\(x^3+3x^2-4\)

\(=x^3-x^2+4x^2-4\)

\(=x^2\left(x-1\right)+4\left(x^2-1\right)\)

\(=x^2\left(x-1\right)+4\left(x+1\right)\left(x-1\right)\)

\(=\left(x-1\right)\left[x^2+4\left(x+1\right)\right]\)

\(=\left(x-1\right)\left(x^2+4x+4\right)\)

\(=\left(x-1\right)\left(x+2\right)^2\)

=.= hok tốt!!

a, \(x^3-x^2-4\)

\(=x^3-2x^2+x^2-2x+2x-4\)

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

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

a) \(x^3-x^2-4\)

\(=x^3-2x^2+x^2-2x+2x-4\)

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

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

b) \(x^8-98x^4+1\)

\(=\left(x^4\right)^2+2\cdot x^4\cdot1+1^2-100x^4\)

\(=\left(x^4+1\right)^2-\left(10x^2\right)^2\)

\(=\left(x^4-10x^2+1\right)\left(x^4+10x^2+1\right)\)

P = -4 -x^2 + 6x

P = - (x^2 -6x+4)

P = - (x^2 - 2.3.x + 9 - 5)

P = -[(x-3)^2 - 5]

P = - (x-3)^2 + 5 =< 5

Để P GTLN

=> -(x-3)^2 + 5 = 5

-(x-3)^2 = 0

x- 3 = 0

x = 3

=> GTLN của P = 5 tại x = 3

Ta có: \(P=-4-x^2+6x=-x^2+6x-4=-\left(x^2-6x+4\right)\)

\(=-\left(x^2-6x+9-5\right)\)

\(=-\left(x^2-6x+9\right)+5\)

\(=-\left(x-3\right)^2+5\le5\forall x\)

Dấu = xảy ra khi: \(-\left(x-3\right)^2=0\Rightarrow x=3\)

Vậy GTLN của P là 5 tại x = 3

=.= hok tốt!!

chao toi la nguoi moi

wow

\(x^3+6x^2+5x=x\left(x^2+6x+5\right)=x\left(x^2+x+5x+5\right)=x\left[x\left(x+1\right)+5\left(x+1\right)\right]\)

\(=x\left(x+1\right)\left(x+5\right)\)