Giải bài 2.33 trang 52 SGK Toán 8 – Cùng khám phá

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)   

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