Đề bài

Tính:

a) \(\dfrac{{3{a^2}}}{{10{b^3}}} \cdot \dfrac{{15b}}{{9{a^4}}}\)                                                            b) \(\dfrac{{x - 3}}{{{x^2}}} \cdot \dfrac{{4x}}{{{x^2} - 9}}\)

c) \(\dfrac{{{a^2} - 6a + 9}}{{{a^2} + 3a}} \cdot \dfrac{{2a + 6}}{{a - 3}}\)                                             d) \(\dfrac{{x + 1}}{x} \cdot \left( {x + \dfrac{{2 - {x^2}}}{{{x^2} - 1}}} \right)\)

Phương pháp giải

Tìm ĐKXĐ

Sử dụng quy tắc nhân đa hai phân thức

Lời giải của GV Loigiaihay.com

a) ĐKXĐ: \(a,b \ne 0\)

\(\dfrac{{3{a^2}}}{{10{b^3}}} \cdot \dfrac{{15b}}{{9{a^4}}}\) \( = \dfrac{{3{a^2}.15b}}{{10{b^3}.9{a^4}}} = \dfrac{{45{a^2}b}}{{90{a^4}{b^3}}} = \dfrac{1}{{2{a^2}{b^2}}}\)

b) ĐKXĐ: \(x \ne 0;\;x \ne  \pm 3\)

\(\dfrac{{x - 3}}{{{x^2}}} \cdot \dfrac{{4x}}{{{x^2} - 9}}\) \( = \dfrac{{\left( {x - 3} \right).4x}}{{{x^2}.\left( {{x^2} - 9} \right)}} = \dfrac{{\left( {x - 3} \right).4x}}{{{x^2}\left( {x - 3} \right)\left( {x + 3} \right)}} = \dfrac{4}{{x\left( {x + 3} \right)}}\)

c) ĐKXĐ: \(x \ne 0;x \ne  \pm 3\)

\(\dfrac{{{a^2} - 6a + 9}}{{{a^2} + 3a}} \cdot \dfrac{{2a + 6}}{{a - 3}}\) \( = \dfrac{{{{\left( {a - 3} \right)}^2}.2.\left( {a + 3} \right)}}{{a.\left( {a + 3} \right).\left( {a - 3} \right)}} = \dfrac{{2\left( {a - 3} \right)}}{a}\)

d)  ĐKXĐ: \(x \ne 0;x \ne 1\)

\(\dfrac{{x + 1}}{x} \cdot \left( {x + \dfrac{{2 - {x^2}}}{{{x^2} - 1}}} \right)\) \( = \dfrac{{x + 1}}{x} \cdot \left[ {\dfrac{{x\left( {{x^2} - 1} \right)}}{{{x^2} - 1}} + \dfrac{{2 - {x^2}}}{{{x^2} - 1}}} \right] = \dfrac{{x + 1}}{x} \cdot \left[ {\dfrac{{{x^3} - x + 2 - {x^2}}}{{{x^2} - 1}}} \right]\) \( = \dfrac{{x + 1}}{x} \cdot \dfrac{{{x^3} - {x^2} - x + 2}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\)\( = \dfrac{{{x^3} - {x^2} - x + 2}}{{x\left( {x - 1} \right)}}\)