Dùng tính chất cơ bản của phân thức, chứng minh \(\frac{{{x^4} - 1}}{{x - 1}} = {x^3} + {x^2} + x + 1\).

Ta có \(\frac{{4x - 4}}{{2x\left( {x + 3} \right)}} = \frac{{2\left( {2x - 2} \right)}}{{2x\left( {x + 3} \right)}} = \frac{{2x - 2}}{{x\left( {x + 3} \right)}}\).

Mẫu thức chung: 3x(x + 3)(x + 1).

Ta có:

3x(x + 3)(x + 1) : x(x + 3) = 3(x + 1)

3x(x + 3)(x + 1) : 3x(x + 1) = (x + 3)

Quy đồng mẫu thức ta có:

\(\frac{{2x - 2}}{{x\left( {x + 3} \right)}} = \frac{{\left( {2x - 2} \right).3\left( {x + 1} \right)}}{{x\left( {x + 3} \right).3\left( {x + 1} \right)}} = \frac{{3\left( {2x - 2} \right)\left( {x + 1} \right)}}{{3x\left( {x + 3} \right)\left( {x + 1} \right)}}\)

\( = \frac{{6\left( {x - 1} \right)\left( {x + 1} \right)}}{{3x\left( {x + 3} \right)\left( {x + 1} \right)}} = \frac{{6\left( {{x^2} - 1} \right)}}{{3x\left( {x + 3} \right)\left( {x + 1} \right)}}\);

\[\frac{{x - 3}}{{3x\left( {x + 1} \right)}} = \frac{{\left( {x - 3} \right).\left( {x + 3} \right)}}{{3x\left( {x + 1} \right).\left( {x + 3} \right)}} = \frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{3x\left( {x + 3} \right)\left( {x + 1} \right)}} = \frac{{{x^2} - 9}}{{3x\left( {x + 3} \right)\left( {x + 1} \right)}}\].

Mẫu thức chung: (1 – x)(x + 1)(x² + 1) = (1 – x²)(x² + 1) = 1 – x⁴.

Quy đồng mẫu thức ta có:

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

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

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