Giải bài 2. 33 trang 52 SGK Toán 8 - Cùng khám phá — Không quảng cáo

Đề bài

Rút gọn các biểu thức sau:

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

b)    $\left( {\frac{{{x^2}}}{{{y^2}}} + \frac{y}{x}} \right):\left( {\frac{x}{{{y^2}}} - \frac{1}{y} + \frac{1}{x}} \right)$

Lời giải chi tiết

a)

$\begin{array}{l}\left( {b - \frac{{{a^2} + {b^2}}}{{a + b}}} \right).\left( {\frac{{2b}}{a} - \frac{{4b}}{{a - b}}} \right)\\ = \left( {\frac{{b\left( {a + b} \right) - {a^2} - {b^2}}}{{a + b}}} \right).\left( {\frac{{2b\left( {a - b} \right) - 4ab}}{{a\left( {a - b} \right)}}} \right)\\ = \left( {\frac{{ab + {b^2} - {a^2} - {b^2}}}{{a + b}}} \right).\left( {\frac{{2ab - 2{b^2} - 4ab}}{{a\left( {a - b} \right)}}} \right)\\ = \left( {\frac{{ab - {a^2}}}{{a + b}}} \right).\left( {\frac{{2{b^2} - 2ab}}{{a\left( {a - b} \right)}}} \right)\\ = \left( {\frac{{ab - {a^2}}}{{a + b}}} \right).\left( {\frac{{ - 2b\left( {a - b} \right)}}{{a\left( {a - b} \right)}}} \right)\\ = \left( {\frac{{ab - {a^2}}}{{a + b}}} \right).\frac{{ - 2b}}{a}\\ = \frac{{\left( {ab - {a^2}} \right). - 2b}}{{a\left( {a + b} \right)}}\\ = \frac{{ - 2a{b^2} + 2{a^2}b}}{{{a^2} + ab}}\\ = \frac{{2ab - 2b}}{{a + b}}\end{array}$

b)

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