Quy đồng mẫu thức hai phân thức:

a) $\frac{3x}{2x+4}$ và $\frac{x+3}{x^2-4}$;                                   b) $\frac{x+5}{x^2+4x+4}$ và $\frac{x}{3x+6}$.

Giải

a) Ta có

• $2x+4=2(x+2)$
• $x^2-4=\left(x-2\right)\left(x+2\right)$
• MTC=$2\left(x-2\right)\left(x+2\right)=2\left(x^2-4\right)$
• Quy đồng
  • $\frac{3x}{2x+4}=\frac{3x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}=\frac{3x\left(x-2\right)}{2\left(x^2-4\right)}$
  • $\frac{x+3}{x^2-4}=\frac{2\left(x+3\right)}{2\left(x-2\right)\left(x+2\right)}=\frac{2\left(x+3\right)}{2\left(x^2-4\right)}$

b) Ta có

• $x^2+4x+4={\left(x+2\right)}^2$
• $3x+6=3(x+2)$
• MTC=$3{(x+2)}^2$
• Quy đồng
  • $\frac{x+5}{x^2+4x+4}=\frac{\left(x+5\right).3}{{\left(x+2\right)}^2.3}=\frac{3\left(x+5\right)}{3{\left(x+2\right)}^2}$
  • $\frac{x}{3x+6}=\frac{x\left(x+2\right)}{3(x+2)\left(x+2\right)}=\frac{x\left(x+2\right)}{3{(x+2)}^2}$