Quy đồng mẫu thức các phân thức sau:

a) $\frac{1}{x+2}, \frac{8}{2x-x^2};$

b) $x^2+1, \frac{x^4}{x^2-1};$

c) $\frac{x^3}{x^3-3x^{2}y+3x{y}^2-y^3},\frac{x}{y^2-xy}.$

Giải

a) MTC$=x(2-x)(2+x)$

• $\frac{1}{x+2}=\frac{1}{2+x}=\frac{x\left(2-x\right)}{x\left(2-x\right)\left(2+x\right)}$
• $\frac{8}{2x-x^2}=\frac{8}{x\left(2-x\right)}=\frac{8\left(2+x\right)}{x\left(2-x\right)\left(2+x\right)}$

b) MTC=$x^2-1$

• $x^2+1=\frac{x^2+1}{1}=\frac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}$
• $\frac{x^4}{x^2-1}=\frac{x^4}{x^2-1}$

c) Ta có

• $x^3-3x^{2}y+3x{y}^2-y^3= {\left(x-y\right)}^3$
• $y^2-xy=y(y-x)=-y(x-y)$
• MTC=$y{(x-y)}^3$
• Quy đồng:

     $\frac{x^3}{x^3-3x^{2}y+3x{y}^2-y^3}=\frac{x^3}{{\left(x-y\right)}^3}=\frac{x^{3}y}{y{\left(x-y\right)}^3}$

     $\frac{x}{y^2-xy}=\frac{x}{-y(x-y)}=\frac{-x}{y(x-y)}=\frac{-x{(x-y)}^2}{y{(x-y)}^3}$