\(a.\frac{x^3+6x^2+2x-3}{x^2+5x-3}=\frac{\left(x+1\right)\left(x^2+5x-3\right)}{x^2+5x-3}=x+1\)

\(b.\frac{x^3-3x^2+x-3}{x-3}=\frac{\left(x-3\right)\left(x^2+1\right)}{x-3}=x^2+1\)

\(c.\frac{x^2+3x-10}{x-2}=\frac{\left(x-2\right)\left(x+5\right)}{x-2}=x+5\)

Bài 1 : 

a, Ta có : \(\frac{AB}{BM}=\frac{AC}{CN}=\frac{6}{3}=\frac{8}{4}=2\)cm 

=> BC // MN ( Ta lét đảo ) 

b, *  Ta có : \(\frac{NF}{DN}=\frac{FK}{EK}=\frac{15}{5}=\frac{21}{7}=3\)cm 

=> NK // DE ( Ta lét đảo )

* Ta có : \(\frac{DM}{ME}=\frac{DN}{NF}=\frac{3}{9}=\frac{5}{15}=\frac{1}{3}\)cm 

=> MN // EF ( Ta lét đảo ) 

Bài 2 : mình tam gọi điểm nằm giữa A;B là T nhé, do thiếu á 

a, Theo định lí Ta lét đảo ta có :

 \(\frac{AT}{AB}=\frac{x}{BC}\Rightarrow\frac{2}{AT+TB}=\frac{x}{6,5}\Rightarrow\frac{2}{5}=\frac{x}{6,5}\Rightarrow x=\frac{13}{5}\)

b, Theo định lí Ta lét đảo ta có : 

\(\frac{DM}{DE}=\frac{x}{ÈF}\Rightarrow\frac{DM}{DM+ME}=\frac{x}{12}\Rightarrow\frac{15}{15+9}=\frac{x}{12}\Rightarrow x=\frac{15}{2}\)

ko biết

\(a)\)

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

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

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

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

\(b)\)

\(x^3+2x^2-6x-27\)

\(=x^3+5x^2+9x-3x^2-15x-27\)

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

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

\(c)\)

\(12x^3+4x^2-27x-9\)

\(=\left(12x^3+4x^2\right)-\left(27x+9\right)\)

\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)

\(=\left(3x+1\right)\left(4x^2-9\right)\)

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

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

\(d)\)

\(16x^2+4x-y^2+y^2\)

\(=16x^2+4x\)

\(4x\left(4x+1\right)\)