a)\(\overrightarrow{AB}+\overrightarrow{AN}=\overrightarrow{AM}\)

b)\(\overrightarrow{BA}+\overrightarrow{CN}=2\overrightarrow{BA}\)

c)Có \(\overrightarrow{AB}=\overrightarrow{NC}\)=>bt trở thành \(\overrightarrow{NC}+MC+\overrightarrow{MN}=\overrightarrow{MC}-\overrightarrow{MC}=vt0\)

d)có vt BA+vt BC=vtBN

bt trở thành vtMN-vtMN=vt0

hok tốt!

g)\(x^m\left(x^2-1\right)\)

h)\(\left(a+b-c\right)\left(x+1\right)\)

i)\(x\left(a+b+c\right)\left(x-1\right)\)

k)\(\left(8x+9\right)\left(7x-5y\right)\)

a) \(\left(a-b\right)\left(a-b-b+a\right)=\left(a-b\right)\left(2a\right)\)

b)\(\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)=\left(a+b\right)\left(-b^2\right)\)

c)\(\left(a+b\right)\left(1-x+x^2\right)\)

d)\(\left(a-b\right)\left(x^2+x-1\right)\)

e)\(\left(a+b\right)\left(c+1\right)\)

ko có j

4)\(\dfrac{2x^3-x+5x^2}{x+1}=\dfrac{\left(x+1\right)\left(2x^2+3x-4\right)+4}{x+1}\)

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

-.-

3)\(\dfrac{2x^4-x^3-3x^2-2x}{x-2}=\dfrac{\left(x-2\right)\left(2x^3+3x^2+3x+4\right)-8}{x-2}=\)\(\left(2x^3+3x^2+3x+4\right)\)\(-\dfrac{8}{x-2}\)

Dùng lược đồ Hóc-ne chia ra đấy bn

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

5)\(\dfrac{x^3-2x^2-9x+18}{x-3}=\dfrac{\left(x-2\right)\left(x-3\right)\left(x+3\right)}{x-3}=\left(x+3\right)\left(x-2\right)=x^2+x-6\)