Giải bài 2.20 trang 50 SGK Toán 8 – Cùng khám phá

Đề bài

Thực hiện các phép tính sau:

a)     \(\frac{{5a}}{{9b}}.\frac{{2a{c^2}}}{b}:\frac{{{c^3}}}{{8{b^3}}}\)

b)    \(\frac{{{x^2} – 2xy}}{{x – y}}.\frac{{y – x}}{{3x – {x^2}}}:\frac{1}{{3 – x}}\)

c)     \(\left( {\frac{{3x}}{{x + 1}} + 1} \right):\left( {1 – \frac{{15{x^2}}}{{1 – {x^2}}}} \right)\)

d)    \(\left( {{m^2} – 1} \right).\left( {\frac{1}{{m + 1}} – \frac{1}{{m – 1}} + 1} \right)\)

Lời giải chi tiết

a)     \(\frac{{5a}}{{9b}}.\frac{{2a{c^2}}}{b}:\frac{{{c^3}}}{{8{b^3}}} = \frac{{5a}}{{9b}}.\frac{{2a{c^2}}}{b}.\frac{{8{b^2}}}{{{c^3}}} = \frac{{5a.2a{c^2}.8{b^2}}}{{9b.b.{c^3}}} = \frac{{5a.2a.8}}{{9.c}} = \frac{{80{a^2}}}{{9c}}\)

b)    \(\begin{array}{l}\frac{{{x^2} – 2xy}}{{x – y}}.\frac{{y – x}}{{3x – {x^2}}}:\frac{1}{{3 – x}} = \frac{{{x^2} – 2xy}}{{x – y}}.\frac{{y – x}}{{3x – {x^2}}}.\frac{{3 – x}}{1} = \frac{{x\left( {x – 2y} \right). – \left( {x – y} \right).\left( {3 – x} \right)}}{{\left( {x – y} \right).x\left( {3 – x} \right)}}\\ =  – \left( {x – 2y} \right)\end{array}\)

c)     \(\begin{array}{l}\left( {\frac{{3x}}{{x + 1}} + 1} \right):\left( {1 – \frac{{15{x^2}}}{{1 – {x^2}}}} \right) = \left( {\frac{{3x}}{{x + 1}} + \frac{{x + 1}}{{x + 1}}} \right):\left( {\frac{{1 – {x^2}}}{{1 – {x^2}}} – \frac{{15{x^2}}}{{1 – {x^2}}}} \right)\\ = \left( {\frac{{4x + 1}}{{x + 1}}} \right):\left( {\frac{{1 – 16{x^2}}}{{1 – {x^2}}}} \right) = \frac{{4x + 1}}{{x + 1}}.\frac{{1 – {x^2}}}{{1 – 16{x^2}}} = \frac{{ – \left( {1 – 4x} \right).\left( {1 – x} \right)\left( {1 + x} \right)}}{{ – \left( {1 – x} \right).\left( {1 – 4x} \right)\left( {1 + 4x} \right)}} = \frac{{1 + x}}{{1 + 4x}}\end{array}\)

d)    \(\begin{array}{l}\left( {{m^2} – 1} \right).\left( {\frac{1}{{m + 1}} – \frac{1}{{m – 1}} + 1} \right) = \left( {{m^2} – 1} \right).\left( {\frac{{m – 1 – m – 1 + \left( {m + 1} \right).\left( {m – 1} \right)}}{{{m^2} – 1}}} \right)\\ = \left( {{m^2} – 1} \right).\left( {\frac{{ – 2 + {m^2} – 1}}{{{m^2} – 1}}} \right) = \left( {{m^2} – 1} \right).\left( {\frac{{ – 3 + {m^2}}}{{{m^2} – 1}}} \right) =  – 3 + {m^2}\end{array}\)