\(\text{a) }x^2+4x+3\\ =x^2+x+3x+3\\ =\left(x^2+x\right)+\left(3x+3\right)\\ =x\left(x+1\right)+3\left(x+1\right)\\ =\left(x+3\right)\left(x+1\right)\\ \)

\(\text{b) }x^2-13x+12\\ =x^2-x-12x+12\\ =\left(x^2-x\right)-\left(12x-12\right)\\ =x\left(x-1\right)-12\left(x-1\right)\\ =\left(x-12\right)\left(x-1\right)\\ \)

\(\text{c) }x^2+5x-6\\ =x^2-x+6x-6\\ =\left(x^2-x\right)+\left(6x-6\right)\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x-1\right)\left(x+1\right)\\ \)

\(\text{d) }2x^2+3x-5\\ =2x^2-2x+5x-5\\ =\left(2x^2-2x\right)+\left(5x-5\right)\\ =2x\left(x-1\right)+5\left(x-1\right)\\ =\left(2x+5\right)\left(x-1\right)\\ \)

\(\text{e) }a^{m+3}-a^m+a-1\\ =\left(a^{m+3}-a^m\right)+\left(a-1\right)\\ =a^m\left(a^3-1\right)+\left(a-1\right)\\ =a^m\left(a-1\right)\left(a^2+a+1\right)+\left(a-1\right)\\ =\left(a-1\right)\left[a^m\left(a^2+a+1\right)+1\right]\\ =\left(a-1\right)\left(a^{m+2}+a^{m+1}+1\right)\\ \)

\(\text{a) }6x-9-x^2\\ =-\left(x^2+6x+9\right)\\ =-\left(x+3\right)^2\) \(\text{b) }5x^3-10x^2y+5xy^2\\ =5x\left(x^2-2xy+y^2\right)\\ =5x\left(x-y\right)^2\)

\(\text{c) }5x^2-10xy+5y^2-20z^2\\ =5\left(x^2-2xy+y^2-4z^2\right)\\ =5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\\ =5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\\ =5\left(x-y-2z\right)\left(x-y+2z\right)\) \(\text{d) }x^2+5x+6\\ =x^2+3x+2x+6\\ =\left(x^2+3x\right)+\left(2x+6\right)\\ =x\left(x+3\right)+2\left(x+3\right)\\ =\left(x+2\right)\left(x+3\right)\)

\(\text{e) }a^2+9a+8\\ =a^2+a+8a+8\\ =\left(a^2+a\right)+\left(8a+8\right)\\ =a\left(a+1\right)+8\left(a+1\right)\\ =\left(a+8\right)\left(a+1\right)\) \(\text{f) }x^2-x-6\\ =x^2-3x+2x-6\\ =\left(x^2-3x\right)+\left(2x-6\right)\\ =x\left(x-3\right)+2\left(x-3\right)\\ =\left(x+2\right)\left(x-3\right)\)

\(\text{g) }x^2+6x+5\\ =x^2+x+5x+5\\ =\left(x^2+x\right)+\left(5x+5\right)\\ =x\left(x+1\right)+5\left(x+1\right)\\ =\left(x+1\right)\left(x+5\right)\)

\(\text{h) }x^3-3x+2\\=x^3-4x+x+2-2x^2+2x^2+2\\ =\left(x^3-2x^2+x\right)-\left(2x^2-4x+2\right)\\ = x\left(x^2-2x+1\right)-2\left(x^2-2x+1\right)\\ =\left(x-2\right)\left(x^2-2x+1\right)\\ =\left(x-2\right)\left(x-1\right)^2\)

\(A=-x^2+4x-5\\ A=-\left(x^2-4x+4\right)-1\\ A=-\left(x^2-2\cdot x\cdot2+2^2\right)-1\\ A=-\left(x-2\right)^2-1\\ Do\text{ }\left(x-2\right)^2\ge0\forall x\\ \Leftrightarrow-\left(x-2\right)^2\le0\forall x\\ \Leftrightarrow A=-\left(x-2\right)^2-1\le-1\forall x\\ \text{Dấu }"="\text{ xảy ra khi: }\\ \left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \text{Vậy }A_{\left(Max\right)}=-1\text{ }khi\text{ }x=2\)

\(B=-2x^2-6x+5\\ B=-2x^2-6x-\dfrac{9}{2}+\dfrac{19}{2}\\ B=-\left(2x^2+6x+\dfrac{9}{2}\right)+\dfrac{19}{2}\\ B=-2\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{19}{2}\\ B=-2\left[x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2\right]+\dfrac{19}{2}\\ B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\\ Do\text{ }\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow2\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\\ \Leftrightarrow-2\left(x+\dfrac{3}{2}\right)^2\le0\forall x\\ \Leftrightarrow B=-2\left(x+\dfrac{3}{2}\right)^2+\dfrac{19}{2}\le\dfrac{19}{2}\forall x\\ \text{Dấu }"="\text{ xảy }ra\text{ }khi:\\ \left(x+\dfrac{3}{2}\right)^2=0\\ \Leftrightarrow x+\dfrac{3}{2}=0\\ \Leftrightarrow x=-\dfrac{3}{2}\\ \text{Vậy }B_{\left(Max\right)}=\dfrac{19}{2}\text{ }khi\text{ }x=-\dfrac{3}{2}\)

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D. Bạn và tôi cùng đến trường.