Giải bài 2.19 trang 50 SGK Toán 8 - Cùng khám phá

Đề bài

Thực hiện các phép tính sau:

a)     \(\frac{{15{a^2}}}{{8bc}}.\frac{{4c}}{{5a{b^2}}}\)

b)    \(\frac{{14{x^3}}}{{5y{z^3}}}:\frac{{7x}}{{15y{z^2}}}\)

c)     \(\frac{{6t + 12}}{{10 - 5t}}.\frac{{t - 2}}{{t + 2}}\)

d)    \(\frac{{m - 5}}{{{m^2} + 1}}:\left( {3m - 15} \right)\)

Lời giải chi tiết

a)     \(\frac{{15{a^2}}}{{8bc}}.\frac{{4c}}{{5a{b^2}}} = \frac{{15{a^2}.4c}}{{8bc.5a{b^2}}} = \frac{{3a.1}}{{2b.{b^2}}} = \frac{{3a}}{{2{b^3}}}\)

b)    \(\frac{{14{x^3}}}{{5y{z^3}}}:\frac{{7x}}{{15y{z^2}}} = \frac{{14{x^3}}}{{5y{z^3}}}.\frac{{15y{z^2}}}{{7x}} = \frac{{14{x^3}.15y{z^2}}}{{5y{z^3}.7x}} = \frac{{2{x^2}.3}}{z} = \frac{{6{x^2}}}{z}\)

c)     \(\frac{{6t + 12}}{{10 - 5t}}.\frac{{t - 2}}{{t + 2}} = \frac{{\left( {6t + 12} \right).\left( {t - 2} \right)}}{{\left( {10 - 5t} \right).\left( {t + 2} \right)}} = \frac{{6\left( {t + 2} \right).\left( {t - 2} \right)}}{{5\left( {2 - t} \right)\left( {t + 2} \right)}} = \frac{6}{{ - 5}}\)

d)    \(\frac{{m - 5}}{{{m^2} + 1}}:\left( {3m - 15} \right) = \frac{{m - 5}}{{{m^2} + 1}}.\frac{1}{{3m - 15}} = \frac{{\left( {m - 5} \right)}}{{\left( {{m^2} + 1} \right).3\left( {m - 5} \right)}} = \frac{1}{{3\left( {{m^2} + 1} \right)}}\)