Bài 3:
\(a,A=\dfrac{2x\left(x+2\right)\left(x-2\right)^2}{x\left(x^2-4\right)\left(x+1\right)}=\dfrac{2\left(x+2\right)\left(x-2\right)^2}{\left(x+1\right)\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{x+1}\\ b,B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)
Bài 4:
\(a,=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\\ b,=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a-b+c\right)}=\dfrac{a+b-c}{a-b+c}\\ c,=\dfrac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}=\dfrac{2x^2-6x+5x-15}{3x^2-9x-x+3}\\ =\dfrac{\left(x-3\right)\left(2x+5\right)}{\left(x-3\right)\left(3x-1\right)}=\dfrac{2x+5}{3x-1}\)
