Giải chi tiết TH 1 Bài 7. Nhân – chia phân thức (trang 36, 37) – SGK Toán 8 Chân trời sáng tạo. Hướng dẫn: Tìm ĐKXĐ.
Câu hỏi/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) \(\dfrac{{x + 1}}{x} \cdot \left( {x + \dfrac{{2 – {x^2}}}{{{x^2} – 1}}} \right)\)
Hướng dẫn:
Tìm ĐKXĐ
Sử dụng quy tắc nhân đa hai phân thức
Lời giải:
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: \(x \ne 0;x \ne 1\)
\(\dfrac{{x + 1}}{x} \cdot \left( {x + \dfrac{{2 – {x^2}}}{{{x^2} – 1}}} \right)\) \( = \dfrac{{x + 1}}{x} \cdot \left[ {\dfrac{{x\left( {{x^2} – 1} \right)}}{{{x^2} – 1}} + \dfrac{{2 – {x^2}}}{{{x^2} – 1}}} \right] = \dfrac{{x + 1}}{x} \cdot \left[ {\dfrac{{{x^3} – x + 2 – {x^2}}}{{{x^2} – 1}}} \right]\) \( = \dfrac{{x + 1}}{x} \cdot \dfrac{{{x^3} – {x^2} – x + 2}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\)\( = \dfrac{{{x^3} – {x^2} – x + 2}}{{x\left( {x – 1} \right)}}\)