Biết \[\frac{{{\rm{x + 3}}}}{{{{\rm{x}}^{\rm{2}}} - {\rm{4}}}}{\rm{.}}\frac{{{\rm{8}} - {\rm{12x + 6}}{{\rm{x}}^{\rm{2}}} - {{\rm{x}}^{\rm{3}}}}}{{{\rm{9x + 27}}}}{\rm{ = }} - \frac{{\rm{A}}}{{\rm{B}}}\]. Tìm A, B.

Lời giải

Đáp án đúng là: A

Do \[a + b + c = 0\] nên \[a = - \left( {b + c} \right)\].

• \[4bc - {a^2} = 4bc - {\left[ { - \left( {b + c} \right)} \right]^2} = 4bc - \left( {{b^2} + 2bc + {c^2}} \right)\]

\[ = 2bc - {b^2} - {c^2} = - {\left( {b - c} \right)^2}\]

• \[bc + 2{a^2} = {a^2} + bc + {a^2} = {a^2} + bc + a\left[ { - \left( {b + c} \right)} \right]\]

\[ = {a^2} + bc - ab - ac\]\( = \left( {{a^2} - ab} \right) - \left( {ac - bc} \right)\)

\( = a\left( {a - b} \right) - c\left( {a - b} \right) = \left( {a - c} \right)\left( {a - b} \right)\)

Khi đó \(\frac{{4bc - {a^2}}}{{bc + 2{a^2}}} = \frac{{ - {{\left( {b - c} \right)}^2}}}{{\left( {a - c} \right)\left( {a - b} \right)}}\).

Tương tự, ta có: \(\frac{{4ca - {b^2}}}{{ca + 2{b^2}}} = \frac{{ - {{\left( {c - a} \right)}^2}}}{{\left( {b - a} \right)\left( {b - c} \right)}};\)

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

\(A = \frac{{4bc - {a^2}}}{{bc + 2{a^2}}} \cdot \frac{{4ca - {b^2}}}{{ca + 2{b^2}}} \cdot \frac{{4ab - {c^2}}}{{ab + 2{c^2}}}\)

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

Lời giải

Đáp án đúng là: D

\[\frac{{3{\rm{x}} + 12}}{{4{\rm{x}} - 16}} \cdot \frac{{8 - 2{\rm{x}}}}{{{\rm{x}} + 4}} = \frac{{3\left( {{\rm{x}} + 4} \right)}}{{4\left( {{\rm{x}} - 4} \right)}} \cdot \frac{{2\left( {4 - {\rm{x}}} \right)}}{{{\rm{x}} + 4}}\]

\[ = \frac{{3\left( {{\rm{x}} + 4} \right)}}{{4\left( {{\rm{x}} - 4} \right)}} \cdot \frac{{ - 2\left( {{\rm{x}} - 4} \right)}}{{{\rm{x}} + 4}} = \frac{{ - 3}}{2}\].