Dựng hình bình hành ABCD.

 Khi đó vì M là trung điểm AC nên M cũng là trung điểm BD \(\Rightarrow\) M, B, D thẳng hàng.

 Lại có ABCD là hình bình hành \(\Rightarrow AD=BC\)

 Mà \(AK=BC\) nên \(AK=AD\)

 \(\Rightarrow\Delta ADK\) cân tại A

 \(\Rightarrow\widehat{AKD}=\widehat{ADK}\) hay \(\widehat{LKB}=\widehat{KDA}\)

 Mặt khác, LB//AD nên \(\widehat{KDA}=\widehat{KBL}\) (2 góc đồng vị)

 Do đó \(\widehat{LKB}=\widehat{KBL}\left(=\widehat{KDA}\right)\)

 \(\Rightarrow\Delta KBL\) cân tại L

 \(\Rightarrow LK=BL\) (đpcm)

\(\dfrac{P\left(x\right)}{x-3}=\dfrac{x^3-2x^2-x+a}{x-3}\)

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

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

Để P(x) chia x-3 dư 5 thì a+6=5

=>a=-1

\(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\\ =\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]-4\\ =\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\\ =\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt: `12x^2+11x=t` 

 \(\left(t+2\right)\left(t-1\right)-4=t^2+2t-t-2-4=t^2-t-6\\ =t^2-3t+2t-6\\ =t\left(t-3\right)+2\left(t-3\right)\\ =\left(t-3\right)\left(t+2\right)\\ =\left(12x^2+11x-3\right)\left(12x^2+11x+2\right)\) 

\(A=x^3-5x^2+3x+9\\ =x^3+x^2-6x^2-6x+9x+9\\ =x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\\ =\left(x^2-6x+9\right)\left(x+1\right)\\ =\left(x-3\right)^2\left(x+1\right)\)

A = \(x^3\) - 5\(x^2\) + 3\(x\) + 9

A = \(x^3\) - 3\(x^2\) - \(x^2\) - \(x^2\) + 3\(x\) + 9

A = (\(x^3\) - 3\(x^2\)) - (\(x^2\) - 3\(x\)) - (\(x^2\) - 9)

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

A = (\(x\) - 3)(\(x^2\) - \(x\) - \(x\) - 3)

A = (\(x\) - 3)[\(x^2\) - (\(x+x\)) - 3]

A = (\(x\) - 3)[\(x^2\) - 2\(x\) - 3]

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

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

\(A=\left(4x^2-2x-1\right)-\left(x^2-4x+2\right)\\ =4x^2-2x-1-x^2+4x-2\\ =3x^2+2x-3\)

Thay `x=-1/2` vào A ta có:

\(A=3\cdot\left(-\dfrac{1}{2}\right)^2+2\cdot-\dfrac{1}{2}-3=\dfrac{3}{8}-1-3=\dfrac{3}{8}-4=-\dfrac{29}{4}\)

\(\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)-24\)

\(=\left(a^2+7a+10\right)\left(a^2+7a+12\right)-24\)

Đặt a^2 + 7a = t 

\(\left(t+10\right)\left(t+12\right)-24=t^2+22t+120-24\)

\(=\left(t+6\right)\left(t+16\right)\)

\(\Rightarrow\left(a^2+7a+6\right)\left(a^2+7a+16\right)=\left(a+1\right)\left(a+6\right)\left(a^2+7a+16\right)\)

a) \(\left(2x-5\right)\left(3x+b\right)=ax^2+bx+c\)

\(\Rightarrow6x^2+2bx-15x-5b=ax^2+bx+c\)

\(\Rightarrow6x^2+\left(2b-15\right)x-5b=ax^2+bx+c\)

\(\Rightarrow\left\{{}\begin{matrix}a=6\\2b-15=b\\-5b=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=6\\b=15\\c=-75\end{matrix}\right.\)

b) \(\left(ax+b\right)\left(x^2-x-1\right)=ax^3+cx^2-1\)

\(\Rightarrow ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1\)

\(\Rightarrow ax^3-ax^2+bx^2-ax-bx-b=ax^3+cx^2-1\)

\(\Rightarrow ax^3+\left(b-a\right)x^2-\left(a+b\right)x-b=ax^3+cx^2-1\)

\(\Rightarrow\left\{{}\begin{matrix}b-a=c\\-\left(a+b\right)=0\\-b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b-a=c\\a=-b\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=1\\c=2\end{matrix}\right.\)

c) \(ax\left(x-4\right)-b\left(x+6\right)+5=2x^2+5x\left(a-b\right)-6x+c\)

\(\Rightarrow ax^2-4ax-bx-6b+5=2x^2+\left(5a-5b\right)x-6x+c\)

\(\Rightarrow ax^2-\left(4a+b\right)x-\left(5a-5b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow ax^2-\left(4a+b+5a-5b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow ax^2-\left(9a-4b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\-\left(9a-4b\right)=-6\\-6b+5=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{9a-6}{4}\\c=-6b+5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=-13\end{matrix}\right.\)