Giải bài 6.6 trang 7 sách bài tập toán 11 - Kết nối tri thức với cuộc sống

Đề bài

Cho \(a\) và \(b\) là hai số dương, \(a \ne b\). Rút gọn biểu thức sau:

\(A = \left[ {\frac{{a - b}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}{b^{\frac{1}{4}}}}} - \frac{{{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}}}{{{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}}}} \right]:\left( {{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}} \right).\)

Lời giải chi tiết

\(\frac{{a - b}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}{b^{\frac{1}{4}}}}} = \frac{{a - b}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}}\)

\( \Rightarrow B = \frac{{a - b}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}{b^{\frac{1}{4}}}}} - \frac{{{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}}}{{{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}}} = \frac{{a - b}}{{{a^{\frac{3}{4}}} + {a^{\frac{1}{2}}}{b^{\frac{1}{4}}}}} - \frac{{{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}}}{{{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}}} = \frac{{a - b}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}} - \frac{{{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}}}{{{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}}}\)

\( = \frac{{a - b - {a^{\frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right)}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}} = \frac{{{a^{\frac{1}{2}}}{b^{\frac{1}{2}}} - b}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}} = \frac{{{b^{\frac{1}{2}}}\left( {{a^{\frac{1}{2}}} - {b^{\frac{1}{2}}}} \right)}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}}.\)

Tacó: \({a^{\frac{1}{2}}} - {b^{\frac{1}{2}}} = \left( {{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}} \right)\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)\)

nên \(B = \frac{{{b^{\frac{1}{2}}} \cdot \left( {{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}} \right)\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}}{{{a^{\frac{1}{2}}}\left( {{a^{\frac{1}{4}}} + {b^{\frac{1}{4}}}} \right)}} = {\left( {\frac{b}{a}} \right)^{\frac{1}{2}}} \cdot \left( {{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}} \right).\)

Do đó\(A = {\left( {\frac{b}{a}} \right)^{\frac{1}{2}}} \cdot \left( {{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}} \right) \cdot \frac{1}{{{a^{\frac{1}{4}}} - {b^{\frac{1}{4}}}}} = {\left( {\frac{b}{a}} \right)^{\frac{1}{2}}}\)