Giải bài 2 trang 21 vở thực hành Toán 8 tập 2

Đề bài

Hãy thực hiện các phép tính đã chỉ ra.

a) $\frac{{4{\rm{x}} - 6}}{{5{{\rm{x}}^2} - x}}.\frac{{25{{\rm{x}}^2} - 10{\rm{x}} + 1}}{{27 + 8{{\rm{x}}^3}}}$;

b) $\frac{{2{\rm{x}} + 10}}{{{{\left( {x - 3} \right)}^2}}}:\frac{{{{\left( {x + 5} \right)}^3}}}{{{x^2} - 9}}$.

Lời giải chi tiết

a) $\frac{{4{\rm{x}} - 6}}{{5{{\rm{x}}^2} - x}}.\frac{{25{{\rm{x}}^2} - 10{\rm{x}} + 1}}{{27 + 8{{\rm{x}}^3}}} = \frac{{\left( {4{\rm{x}} - 6} \right)\left( {25{{\rm{x}}^2} - 10{\rm{x}} + 1} \right)}}{{\left( {5{{\rm{x}}^2} - x} \right)\left( {27 + 8{{\rm{x}}^3}} \right)}}$

$= \frac{{ - 2\left( {3 - 2{\rm{x}}} \right){{\left( {5{\rm{x}} - 1} \right)}^2}}}{{x\left( {5{\rm{x}} - 1} \right)\left( {3 - 2{\rm{x}}} \right)\left( {9 + 6{\rm{x}} + 4{{\rm{x}}^2}} \right)}} = \frac{{ - 2\left( {5{\rm{x}} - 1} \right)}}{{x\left( {9 + 6{\rm{x}} + 4{{\rm{x}}^2}} \right)}}$

b) $\frac{{2{\rm{x}} + 10}}{{{{\left( {x - 3} \right)}^2}}}:\frac{{{{\left( {x + 5} \right)}^3}}}{{{x^2} - 9}} = \frac{{2{\rm{x}} + 10}}{{{{\left( {x - 3} \right)}^2}}}.\frac{{{x^2} - 9}}{{{{\left( {x + 5} \right)}^2}}}$

$= \frac{{2\left( {x + 5} \right)}}{{{{\left( {x - 3} \right)}^2}}}.\frac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{{{\left( {x + 5} \right)}^3}}} = \frac{{2\left( {x + 3} \right)}}{{\left( {x - 3} \right){{\left( {x + 5} \right)}^2}}}$.