\(B=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)

\(B_{max}=-2\) khi \(x=3\)

\(B=-(x-3)^2-2\le2 \)

Đẳng thức xảy ra khi x = 3.

Vậy GTLN của B là 2 (khi x = 3)

Phân tích đa thức thành nhân tử:

a) x² (2x - 5) + 6x - 15 =  x² (2x - 5) + 3(2x - 5) =( x2 + 3 )(2x - 5)

b)x² + 7x + 12 =  x² + 3x + 4x + 12 = (x + 3 )( x + 4)

a: \(x^2-6x+5=\left(x-5\right)\left(x-1\right)\)

b: \(x^2-x-12=\left(x-4\right)\left(x+3\right)\)

c: \(x^2+8x+15=\left(x+5\right)\left(x+3\right)\)

d: \(2x^2-5x-12=\left(x-4\right)\left(2x+3\right)\)

e: \(x^2-13x+36=\left(x-9\right)\left(x-4\right)\)

a: =3x^2-3x-8x+8=(x-1)(3x-8)

b: =x^2-x-5x+5=(x-1)(x-5)

c: =x^2-6x+2x-12=(x-6)(x+2)

\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)

\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)

\(\Leftrightarrow-11x=-22\)

hay x=2

b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)

\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)

\(\Leftrightarrow x=-5\)

8 x 3  + 12 x 2  + 6x + 1 = 0

2 x 3  + 3. 2 x 2 .1 + 3.(2x). 1 2  + 1 3  = 0

2 x + 1 3  = 0

2x + 1 = 0

x = (-1)/2

\(P=\left(x^2-6x+9\right)+\left(y^2-2y+1\right)+2\\ P=\left(x-3\right)^2+\left(y-1\right)^2+2\ge2\\ P_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

a) (x + 2)(x + 4).              b) 2(x + 6)(x + l).

c) 3(3x + 5)(x + l).           d) (6x -7y)(x + y).