Đơn giản và rút gọn biểu thức
a) \({{{t^2} - t - 6} \over {{t^2} - 6t + 9}}\) ;
b) \({{3a{b^3}} \over {8{a^2}b}}.{{4ac} \over {9{b^4}}}\) ;
c) \( - {4 \over {8x}}:{{16} \over {x{y^2}}}\) ;
d) \({{48} \over {6a + 42}}.{{7a + 49} \over {16}}\) ;
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e) \({{{{\rm{w}}^2} + 5{\rm{w + 4}}} \over 6}:{{{\rm{w}} + 1} \over {18w + 24}}\) ;
f) \({{{{{x^2} + x} \over {x + 1}}} \over {{x \over {x - 1}}}}\) .
\(\eqalign{ & a)\,\,{{{t^2} - t - 6} \over {{t^2} - 6t + 9}} = {{\left( {t - 3} \right)\left( {t + 2} \right)} \over {{{\left( {t - 3} \right)}^2}}} = {{t + 2} \over {t - 3}} \cr & b)\,\,{{3a{b^3}} \over {8{a^2}b}}.{{4ac} \over {9{b^4}}} = {{3{b^2}} \over {8a}}.{{4ac} \over {9{b^4}}} = {{12a{b^2}c} \over {72a{b^4}}} = {c \over {6{b^3}}} \cr & c)\,\, - {4 \over {8x}}:{{16} \over {x{y^2}}} = {{ - 1} \over {2x}}.{{x{y^2}} \over {16}} = {{ - {y^2}} \over {32}} \cr & d)\,\,{{48} \over {6a + 42}}.{{7a + 49} \over {16}} = {{48} \over {6\left( {a + 7} \right)}}.{{7\left( {a + 7} \right)} \over {16}} = {{8.7\left( {a + 7} \right)} \over {\left( {a + 7} \right).16}} = {7 \over 2} \cr & e)\,\,{{{w^2} + 5w + 4} \over 6}:{{w + 1} \over {18w + 24}} = {{\left( {w + 4} \right)\left( {w + 1} \right)} \over 6}.{{6\left( {3w + 4} \right)} \over {w + 1}} \cr & = {{\left( {w + 4} \right)\left( {w + 1} \right)6\left( {3w + 4} \right)} \over {6\left( {w + 1} \right)}} = \left( {w + 4} \right)\left( {3w + 4} \right) \cr & f)\,\,{{{{{x^2} + 1} \over {x + 1}}} \over {{x \over {x - 1}}}} = \left( {{{{x^2} + x} \over {x + 1}}} \right):\left( {{x \over {x - 1}}} \right) = {{x\left( {x + 1} \right)} \over {x + 1}}.{{x - 1} \over x} = {x \over 1}.{{x - 1} \over x} = x - 1 \cr} \)