Giải bài 14 trang 39 sách bài tập toán 8 - Cánh diều

Đề bài

Thực hiện phép tính:

a) \(\frac{{24{y^5}}}{{7{x^2}}}.\left( { - \frac{{49x}}{{12{y^3}}}} \right)\)

b) \( - \frac{{36{y^3}}}{{15{x^4}}}.\left( { - \frac{{45{x^2}}}{{9{y^3}}}} \right)\)

c) \(\frac{{{x^2} - {y^2}}}{{{x^2}}}.\frac{{{x^4}}}{{{{\left( {x + y} \right)}^2}}}\)

d) \(\frac{{x + 3}}{{{x^2} - 1}}.\frac{{1 - 3x + 3{x^2} - {x^3}}}{{9x + 27}}\)

Lời giải chi tiết

a) \(\frac{{24{y^5}}}{{7{x^2}}}.\left( { - \frac{{49x}}{{12{y^3}}}} \right) = \frac{{12{y^3}.2{y^2}}}{{7{x^2}}}.\left( { - \frac{{7x.7}}{{12{y^3}}}} \right) =  - \frac{{14{y^2}}}{x}\)

b) \( - \frac{{36{y^3}}}{{15{x^4}}}.\left( { - \frac{{45{x^2}}}{{9{y^3}}}} \right) =  - \frac{{9{y^3}.4}}{{15{x^2}.{x^2}}}.\left( { - \frac{{15{x^2}.3}}{{9{y^3}}}} \right) = \frac{{12}}{{{x^2}}}\)

c) \(\frac{{{x^2} - {y^2}}}{{{x^2}}}.\frac{{{x^4}}}{{{{\left( {x + y} \right)}^2}}} = \frac{{\left( {x + y} \right)\left( {x - y} \right)}}{{{x^2}}}.\frac{{{x^2}.{x^2}}}{{{{\left( {x + y} \right)}^2}}} = \frac{{{x^2}\left( {x - y} \right)}}{{x + y}}\)

d) \(\frac{{x + 3}}{{{x^2} - 1}}.\frac{{1 - 3x + 3{x^2} - {x^3}}}{{9x + 27}} = \frac{{x + 3}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}.\frac{{ - {{\left( {x - 1} \right)}^3}}}{{9\left( {x + 3} \right)}} =  - \frac{{{{\left( {x - 1} \right)}^2}}}{{9\left( {x + 1} \right)}}\)