Tháng Tư 15, 2024

Thực hiện phép tính: a) \(\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\,\,\,:\,\,\frac{{10x – 10y}}{{{x^3} + {y^3}}}\) b) \(\frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3{x^2} – 3x + 3}}{{{x^2} – 36}} + \frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3x}}{{{x^2} – 36}}\) c) \(\frac{{x – 1}}{{{x^2} – 4x + 4}}\, \cdot \,\frac{{{x^2} – 4}}{{{x^3} – 1}}\, \cdot \,\frac{{{x^2} + x + 1}}{{x + 2}}\) d) \(\left( {\frac{{3x}}{{1 – 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 – 6x + 9{x^2}}}\)

Thực hiện phép tính:

a) \(\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\,\,\,:\,\,\frac{{10x – 10y}}{{{x^3} + {y^3}}}\)

b) \(\frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3{x^2} – 3x + 3}}{{{x^2} – 36}} + \frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3x}}{{{x^2} – 36}}\)

c) \(\frac{{x – 1}}{{{x^2} – 4x + 4}}\, \cdot \,\frac{{{x^2} – 4}}{{{x^3} – 1}}\, \cdot \,\frac{{{x^2} + x + 1}}{{x + 2}}\)

d) \(\left( {\frac{{3x}}{{1 – 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 – 6x + 9{x^2}}}\)

A. a) \(\frac{{3({x^2} – {y^2})}}{{50}}.\)

b) \(\frac{1}{x – 2}.\)

c) \(\frac{3}{{x + 6}}.\)

d) \(\frac{{1 – 3x}}{{2(3x + 1)}}.\)

B. a) \(\frac{{3({x} – {y})}}{{50}}.\)

b) \(\frac{1}{x +2}.\)

c) \(\frac{3}{{x – 6}}.\)

d) \(\frac{{3x}}{{2(3x + 1)}}.\)

C. a) \(\frac{{3({x} – {y})}}{{5}}.\)

b) \(\frac{3}{x +2}.\)

c) \(\frac{15}{{x + 6}}.\)

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

D. a) \(\frac{{3({x}^3 – {y}^3)}}{{50}}.\)

b) \(\frac{6}{x +2}.\)

c) \(\frac{3}{{x – 6}^2}.\)

d) \(\frac{{3x}}{{(3x – 1)}}.\)

Hướng dẫn Chọn đáp án là: A

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, 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:

\(a)\,\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\,\,:\,\,\frac{{10x – 10y}}{{{x^3} + {y^3}}}\)

\( = \,\,\frac{{3{x^2} – 6xy + 3{y^2}}}{{5{x^2} – 5xy + 5{y^2}}}\, \cdot \,\,\frac{{{x^3} + {y^3}}}{{10x – 10y}}\)

\( = \frac{{3({x^2} – 2xy + {y^2})}}{{5({x^2} – xy + {y^2})}} \cdot \frac{{(x + y)({x^2} – xy + {y^2})}}{{10(x – y)}}\)

\( = \frac{{3{{(x – y)}^2}}}{{5({x^2} – xy + {y^2})}} \cdot \frac{{(x + y)({x^2} – xy + {y^2})}}{{10(x – y)}} = \frac{{3({x^2} – {y^2})}}{{50}}.\)

\(b)\frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3{x^2} – 3x + 3}}{{{x^2} – 36}} + \frac{{x – 6}}{{{x^2} + 1}}\,\, \cdot \,\,\frac{{3x}}{{{x^2} – 36}}\)

\( = \frac{{x – 6}}{{{x^2} + 1}}\left( {\frac{{3{x^2} – 3x + 3}}{{{x^2} – 36}} + \frac{{3x}}{{{x^2} – 36}}} \right)\)

\( = \frac{{x – 6}}{{{x^2} + 1}} \cdot \frac{{3{x^2} – 3x + 3 + 3x}}{{{x^2} – 36}}\)

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

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

\(\begin{array}{l}c)\frac{{x – 1}}{{{x^2} – 4x + 4}}\, \cdot \,\frac{{{x^2} – 4}}{{{x^3} – 1}}\, \cdot \,\frac{{{x^2} + x + 1}}{{x + 2}}\\= \frac{{x – 1}}{{{{(x – 2)}^2}}}\, \cdot \,\frac{{(x – 2)(x + 2)}}{{(x – 1)({x^2} + x + 1)}}\, \cdot \,\frac{{{x^2} + x + 1}}{{x + 2}}\\= \frac{1}{{x – 2}}.\end{array}\)

\(\begin{array}{l}d)\left( {\frac{{3x}}{{1 – 3x}} + \frac{{2x}}{{3x + 1}}} \right):\frac{{6{x^2} + 10x}}{{1 – 6x + 9{x^2}}}\\= \frac{{3x(3x + 1) + 2x(1 – 3x)}}{{(1 – 3x)(3x + 1)}}:\frac{{2x(3x + 5)}}{{{{(1 – 3x)}^2}}}\\= \frac{{3{x^2} + 5x}}{{(1 – 3x)(3x + 1)}} \cdot \frac{{{{(1 – 3x)}^2}}}{{2x(3x + 5)}}\\= \frac{{x(3x + 5)}}{{(1 – 3x)(3x + 1)}} \cdot \frac{{{{(1 – 3x)}^2}}}{{2x(3x + 5)}} = \frac{{1 – 3x}}{{2(3x + 1)}}.\end{array}\)

Chọn A.