Rút gọn biểu thức: a) \(\left( {\frac{9}{{{x^3} – 9x}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{{x^2} + 3x}} – \frac{x}{{3x + 9}}} \right)\) b) \(\frac{{x – 2}}{{{x^2} + 3x + 2}}\, \cdot \,\frac{{x + 2}}{{{x^2} – 5x + 6}}\) c) \(\frac{{{x^2} + 1}}{{3x}}:\frac{{{x^2} + 1}}{{x – 1}}:\frac{{{x^3} – 1}}{{{x^2} + x}}:\frac{{{x^2} + 2x + 1}}{{{x^2} + x + 1}}\) d) \(\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}}\)

Rút gọn biểu thức:

a) \(\left( {\frac{9}{{{x^3} – 9x}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{{x^2} + 3x}} – \frac{x}{{3x + 9}}} \right)\)

b) \(\frac{{x – 2}}{{{x^2} + 3x + 2}}\, \cdot \,\frac{{x + 2}}{{{x^2} – 5x + 6}}\)

c) \(\frac{{{x^2} + 1}}{{3x}}:\frac{{{x^2} + 1}}{{x – 1}}:\frac{{{x^3} – 1}}{{{x^2} + x}}:\frac{{{x^2} + 2x + 1}}{{{x^2} + x + 1}}\)

d) \(\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}}\)

A. a) \(\frac{{ 3}}{{x – 3}}.\)

b) \(\frac{-1}{{(x + 1)(x – 3)}}.\)

c) \(\frac{1}{{3(x – 1)}}.\)

d) \(\frac{{x – 3}}{{x – 1}}.\)

B. a) \(\frac{{ – 3}}{{x – 3}}.\)

b) \(\frac{1}{{(x + 1)(x – 3)}}.\)

c) \(\frac{1}{{3(x + 1)}}.\)

d) \(\frac{{x + 3}}{{x – 1}}.\)

C. a) \(\frac{{ – 3}}{{x + 3}}.\)

b) \(\frac{1}{{(x -1)(x +3)}}.\)

c) \(\frac{1}{{3(x -1)}}.\)

d) \(\frac{{x -3}}{{x – 1}}.\)

D. a) \(\frac{{ 4}}{{x – 3}}.\)

b) \(\frac{1}{{(x -1)(x – 3)}}.\)

c) \(\frac{-1}{{3(x -1)}}.\)

d) \(\frac{{x + 3}}{{x + 1}}.\)

Hướng dẫn Chọn đáp án là: B

Phương pháp giải:

Áp dụng quy tắc nhân chia hai hay nhiều phân thức, áp dụng tính chất phân phối của phép nhân và phép cộng để tính thuận tiện nhất, thứ tự thực hiện phép tính, rèn luyện kĩ năng phân tích đa thức thành nhân tử, quy đồng, rút gọn.

Lời giải chi tiết:

\(\begin{array}{l}a)\left( {\frac{9}{{{x^3} – 9x}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{{x^2} + 3x}} – \frac{x}{{3x + 9}}} \right)\\= \left( {\frac{9}{{x({x^2} – 9)}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{x(x + 3)}} – \frac{x}{{3(x + 3)}}} \right)\\= \left( {\frac{9}{{x(x – 3)(x + 3)}} + \frac{1}{{x + 3}}} \right):\left( {\frac{{x – 3}}{{x(x + 3)}} – \frac{x}{{3(x + 3)}}} \right)\\= \frac{{9 + x(x – 3)}}{{x(x – 3)(x + 3)}}:\frac{{3(x – 3) – x.x}}{{3x(x + 3)}}\\= \frac{{{x^2} – 3x + 9}}{{x(x – 3)(x + 3)}}:\frac{{ – {x^2} + 3x – 9}}{{3x(x + 3)}}\\= \frac{{{x^2} – 3x + 9}}{{x(x – 3)(x + 3)}} \cdot \frac{{3x(x + 3)}}{{ – ({x^2} – 3x + 9)}}\\= \frac{{ – 3}}{{x – 3}}\end{array}\)

\(\begin{array}{l}b)\frac{{x – 2}}{{{x^2} + 3x + 2}}\, \cdot \,\frac{{x + 2}}{{{x^2} – 5x + 6}}\\= \frac{{x – 2}}{{{x^2} + x + 2x + 2}}\, \cdot \,\frac{{x + 2}}{{{x^2} – 2x – 3x + 6}}\\= \frac{{x – 2}}{{x(x + 1) + 2(x + 1)}}\, \cdot \,\frac{{x + 2}}{{x(x – 2) – 3(x – 2)}}\\= \frac{{x – 2}}{{(x + 1)(x + 2)}}\, \cdot \,\frac{{x + 2}}{{(x – 2)(x – 3)}}\\= \frac{1}{{(x + 1)(x – 3)}}.\end{array}\)

\(c)\frac{{{x^2} + 1}}{{3x}}:\frac{{{x^2} + 1}}{{x – 1}}:\frac{{{x^3} – 1}}{{{x^2} + x}}:\frac{{{x^2} + 2x + 1}}{{{x^2} + x + 1}}\)

\( = \frac{{{x^2} + 1}}{{3x}} \cdot \frac{{x – 1}}{{{x^2} + 1}} \cdot \frac{{{x^2} + x}}{{{x^3} – 1}} \cdot \frac{{{x^2} + x + 1}}{{{x^2} + 2x + 1}}\)

\( = \frac{{x – 1}}{{3x}} \cdot \frac{{x(x + 1)}}{{(x – 1)({x^2} + x + 1)}} \cdot \frac{{{x^2} + x + 1}}{{{x^2} + 2x + 1}}\)

\( = \frac{{x + 1}}{{3({x^2} + x + 1)}} \cdot \frac{{{x^2} + x + 1}}{{{{(x + 1)}^2}}} = \frac{1}{{3(x + 1)}}.\)

\(d)\frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{{x^2} + 6x}} – \frac{{x + 3}}{{{x^2} – 1}}:\frac{{x + 4}}{{x – 4}}\)

\( = \frac{{x + 3}}{{{x^2} – 1}} \cdot \frac{{{x^2} + 6x}}{{x + 4}} – \frac{{x + 3}}{{{x^2} – 1}} \cdot \frac{{x – 4}}{{x + 4}}\)

\( = \frac{{x + 3}}{{{x^2} – 1}} \cdot \left( {\frac{{{x^2} + 6x}}{{x + 4}} – \frac{{x – 4}}{{x + 4}}} \right)\)

\( = \frac{{x + 3}}{{(x – 1)(x + 1)}} \cdot \frac{{{x^2} + 6x – x + 4}}{{x + 4}}\)

\( = \frac{{x + 3}}{{(x – 1)(x + 1)}} \cdot \frac{{{x^2} + 5x + 4}}{{x + 4}}\)

\( = \frac{{x + 3}}{{(x – 1)(x + 1)}} \cdot \frac{{(x + 1)(x + 4)}}{{x + 4}} = \frac{{x + 3}}{{x – 1}}.\)

Chọn B.