Giải mục 1 trang 36, 37 SGK Toán 8 tập 1 – Chân trời sáng tạo

Video hướng dẫn giải

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 chi tiết:

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)}}$