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

b) Ta có: \(\dfrac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)

\(=\dfrac{4^{2m}-4^{2n}}{4^n+4^m}\)

\(=\dfrac{\left(4^m+4^n\right)\left(4^m-4^n\right)}{4^n+4^m}\)

\(=4^m-4^n\)

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

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

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

b: ĐKXĐ: x<>1

\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)

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

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)

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

e: ĐKXĐ: x<>-y

\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

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

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

1) 3a - 3b + a^2 - ab

= 3(a - b) + a(a - b)

= (a - b)(a + 3)

2) = 3xy(x^2 + y^2)/(x^2 + y^2) = 3xy

a: \(A=-2y^3-3y^4+2\)

\(B=7y^4+2y^3+3y^2-3y-3\)

b: \(A+B=4y^4+3y^2-3y-1\)

\(A-B=-10y^4-4y^3-3y^2+3y+5\)

Ta có: \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

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

\(=\dfrac{\left(x^3+y^3\right)\left(x^3+y^3\right)}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}\)

\(=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)

\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}=\dfrac{x^2-xy+y^2}{x^3+xy^2-x^2y-y^3}\)

\(A=\left(\dfrac{4\sqrt{y}}{2+\sqrt{y}}+\dfrac{8y}{4-y}\right):\left(\dfrac{\sqrt{y}-1}{y-2\sqrt{y}}-\dfrac{2}{\sqrt{y}}\right)\)

\(A=\dfrac{4\sqrt{y}\left(2-\sqrt{y}\right)+8y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}:\dfrac{\sqrt{y}-1-2\left(\sqrt{y}-2\right)}{\sqrt{y}\left(\sqrt{y}-2\right)}\)

\(A=\dfrac{8\sqrt{y}+4y}{\left(2+\sqrt{y}\right)\left(2-\sqrt{y}\right)}.\dfrac{\sqrt{y}\left(\sqrt{y}-2\right)}{-\sqrt{y}+3}\)

\(A=\dfrac{4\sqrt{y}}{2-\sqrt{y}}.\dfrac{\sqrt{y}\left(2-\sqrt{y}\right)}{\sqrt{y}-3}\)

\(A=\dfrac{4y}{\sqrt{y}-3}\)

Chúc bạn học tốt ^^