by (1đ) Rút gọn phân thức sau: \(C = \frac{{{a^3} – {b^3} + {c^3} + 3abc}}{{{{(a + b)}^2} + {{(b + c)}^2} + {{(c – a)}^2}}}\).
Lời giải chi tiết:
Ta có:
\(\begin{array}{l}{a^3} – {b^3} + {c^3} + 3abc = ({a^3} + {c^3} + 3{a^2}c + 3a{c^2}) – 3{a^2}c – 3a{c^2} + 3abc – {b^3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {(a + c)^3} – {b^3} – 3ac(a + c – b)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = (a + c – b)\left[ {{{(a + c)}^2} + b(a + c) + {b^2}} \right] – 3ac(a + c – b)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = (a + c – b)({a^2} + {b^2} + {c^2} + ab + bc – ac)\\{(a + b)^2} + {(b + c)^2} + {(c – a)^2} = ({a^2} + 2ab + {b^2}) + ({b^2} + 2bc + {c^2}) + ({c^2} – 2ac + {a^2})\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{a^2} + 2{b^2} + 2{c^2} + 2ab + 2bc – 2ac\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2({a^2} + {b^2} + {c^2} + ab + bc – ac)\\ \Rightarrow C = \frac{{(a + c – b)({a^2} + {b^2} + {c^2} + ab + bc – ac)}}{{2({a^2} + {b^2} + {c^2} + ab + bc – ac)}} = \frac{{a + c – b}}{2}\end{array}\)
