Bài 1.
\({{\left( {x + 1} \right)\left( {{x^2} - 2x + 1} \right)} \over {6{x^3} + 6}}:{{{x^2} - 1} \over {4{x^2} - 4x + 4}} \)
\(\;= {{\left( {x + 1} \right){{\left( {x - 1} \right)}^2}} \over {6\left( {{x^3} + 1} \right)}}:{{\left( {x - 1} \right)\left( {x + 1} \right)} \over {4\left( {{x^2} - x + 1} \right)}}\)
\( \;= {{\left( {x + 1} \right){{\left( {x - 1} \right)}^2}} \over {6\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.{{4\left( {{x^2} - x + 1} \right)} \over {\left( {x - 1} \right)\left( {x + 1} \right)}}\)
\(\;= {{2\left( {x - 1} \right)} \over {3\left( {x - 1} \right)}}.\)
Bài 2.
a) Điều kiện xác định: \(a - 2 \ne 0 \Rightarrow a \ne 2\)
(vì \({a^2} + 2a + 4 = {a^2} + 2a + 1 + 3 \)\(\;= {\left( {a + 1} \right)^2} + 3 > 0\;\forall a)\) .
b) \(A = {{a + 2} \over {a - 2}}\left( {{{6a} \over {{a^3} - 8}} + {{2a} \over {{a^2} + 2a + 4}} + {1 \over {2 - a}}} \right) - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}:\left[ {{{6a} \over {{a^3} - 8}} + {{2a} \over {{a^2} + 2a + 4}} + {1 \over {2 - a}}} \right] - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}:\left[ {{{6a} \over {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)}} + {{2a} \over {{a^2} + 2a + 4}} - {1 \over {a - 2}}} \right] - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}:\left[ {{{6a + 2a\left( {a - 2} \right) - \left( {{a^2} + 2a + 4} \right)} \over {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)}}} \right] - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}:\left[ {{{6a + 2{a^2} - 4a - {a^2} - 2a - 4} \over {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)}}} \right] - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}:{{{a^2} - 4} \over {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)}} - {{4a + 4} \over {a - 2}}\)
\( = {{a + 2} \over {a - 2}}.{{\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)} \over {\left( {a - 2} \right)\left( {a + 2} \right)}} - {{4a + 4} \over {a - 2}} \)
\(= {{{a^2} + 2a + 4} \over {a - 2}} - {{4a + 4} \over {a - 2}}\)
\( = {{{a^2} + 2a + 4 - 4a - 4} \over {a - 2}} \)
\(= {{{a^2} - 2a} \over {a - 2}} = {{a\left( {a - 2} \right)} \over {a - 2}} = a.\)
c) \(a = 2012 \Rightarrow A = 2012\) (thỏa điều kiện xác định).
