Khi quy đồng mẫu thức ba phân thức \(\dfrac{1}{x^2+4x+3};\dfrac{1}{x^2+5x+4};\dfrac{1}{x^2+7x+12}\), ta có kết quả là
\(\dfrac{x+4}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{x+3}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{x+1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\).\(\dfrac{1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\).\(\dfrac{x+4}{\left(x+1\right)\left(x+3\right)};\dfrac{x+3}{\left(x+1\right)\left(x+4\right)};\dfrac{x+1}{\left(x+3\right)\left(x+4\right)}\).\(\dfrac{4}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{3}{\left(x+1\right)\left(x+3\right)\left(x+4\right)};\dfrac{1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\).Hướng dẫn giải:\(\dfrac{1}{x^2+4x+3}=\dfrac{1}{\left(x+1\right)\left(x+3\right)}\); \(\dfrac{1}{x^2+5x+4}=\dfrac{1}{\left(x+1\right)\left(x+4\right)}\); \(\dfrac{1}{x^2+7x+12}=\dfrac{1}{\left(x+3\right)\left(x+4\right)}\).
\(MSC=\left(x+1\right)\left(x+3\right)\left(x+4\right)\).
\(\dfrac{1}{x^2+4x+3}=\dfrac{1}{\left(x+1\right)\left(x+3\right)}=\dfrac{x+4}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\);
\(\dfrac{1}{x^2+5x+4}=\dfrac{1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x+3}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\);
\(\dfrac{1}{x^2+7x+12}=\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{x+1}{\left(x+1\right)\left(x+3\right)\left(x+4\right)}\).
