Giải bài 7 trang 25 sách bài tập toán 8 - Chân trời sáng tạo

Đề bài

Thu gọn các biểu thức sau:

a) $\frac{{16 - {a^2}}}{{{a^2} + 8a + 16}}:\frac{{a - 4}}{{2a + 4}}.\frac{{a + 4}}{{a + 2}}$;

b) $\frac{{{a^2} - ab + {b^2}}}{{{b^2} - {a^2}}}.\frac{{a + b}}{{{a^3} + {b^3}}}:\frac{{a + b}}{{a - b}}$;

c) $\left( {\frac{{2a}}{{a - 2}} - \frac{a}{{a + 2}}} \right).\frac{{{a^2} - 4}}{a}$;

d) $\left( {\frac{1}{{{a^2}}} - \frac{1}{{ab}}} \right).\frac{{a{b^2}}}{{a - b}}$.

Lời giải chi tiết

a) $\frac{{16 - {a^2}}}{{{a^2} + 8a + 16}}:\frac{{a - 4}}{{2a + 4}}.\frac{{a + 4}}{{a + 2}} = \frac{{\left( {4 - a} \right)\left( {a + 4} \right)}}{{{{\left( {a + 4} \right)}^2}}}.\frac{{a + 4}}{{a + 2}}:\frac{{a - 4}}{{2a + 4}}$

$= \frac{{\left( {4 - a} \right)\left( {4 + a} \right)\left( {a + 4} \right)}}{{{{\left( {a + 4} \right)}^2}.\left( {a + 2} \right)}}.\frac{{2\left( {a + 2} \right)}}{{a - 4}} = \frac{{\left( {4 - a} \right)2\left( {a + 2} \right)}}{{\left( {a + 2} \right)\left( {a - 4} \right)}} =  - 2$

b) $\frac{{{a^2} - ab + {b^2}}}{{{b^2} - {a^2}}}.\frac{{a + b}}{{{a^3} + {b^3}}}:\frac{{a + b}}{{a - b}} = \frac{{{a^2} - ab + {b^2}}}{{\left( {b - a} \right)\left( {b + a} \right)}}.\frac{{a + b}}{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}.\frac{{a - b}}{{a + b}}$

$= \frac{{\left( {{a^2} - ab + {b^2}} \right)\left( {a - b} \right)}}{{\left( {b - a} \right)\left( {b + a} \right)\left( {{a^2} - ab + {b^2}} \right)\left( {a + b} \right)}} = \frac{{ - 1}}{{{{\left( {b + a} \right)}^2}}}$

c) $\left( {\frac{{2a}}{{a - 2}} - \frac{a}{{a + 2}}} \right).\frac{{{a^2} - 4}}{a} = \left[ {\frac{{2a\left( {a + 2} \right)}}{{\left( {a - 2} \right)\left( {a + 2} \right)}} - \frac{{a\left( {a - 2} \right)}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}} \right].\frac{{\left( {a - 2} \right)\left( {a + 2} \right)}}{a}$

$= \frac{{2{a^2} + 4a - {a^2} + 2a}}{{\left( {a - 2} \right)\left( {a + 2} \right)}}.\frac{{\left( {a - 2} \right)\left( {a + 2} \right)}}{a} = \frac{{{a^2} + 6a}}{a} = \frac{{a\left( {a + 6} \right)}}{a} = a + 6$

d) $\left( {\frac{1}{{{a^2}}} - \frac{1}{{ab}}} \right).\frac{{a{b^2}}}{{a - b}} = \frac{{b - a}}{{{a^2}b}}.\frac{{a{b^2}}}{{a - b}} = \frac{{ - \left( {a - b} \right)a{b^2}}}{{{a^2}b\left( {a - b} \right)}} = \frac{{ - b}}{a}$.