Rút gọn rồi tính giá trị của biểu thức sau tại \(x = - {1 \over 3}\) :
\(\left[ {{{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {{x^2} - 9}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}} \right]\left[ {1:\left( {{{24{x^2}} \over {{x^4} - 81}} - {{12} \over {{x^2} + 9}}} \right)} \right]\)
Hướng dẫn làm bài:
+Ngoặc vuông thứ nhất:
\(\left[ {{{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {{x^2} - 9}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}} \right]{{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {{x^2} - 9}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}\)
\(= {{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {\left( {x - 3} \right)\left( {x + 3} \right)}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}\left[ {1:\left( {{{24{x^2}} \over {{x^4} - 81}} - {{12} \over {{x^2} + 9}}} \right)} \right]\)
\(={{{{\left( {x + 3} \right)}^2} + 6\left( {x - 3} \right)\left( {x + 3} \right) - {{\left( {x - 3} \right)}^2}} \over {{{\left( {x - 3} \right)}^2}{{\left( {x + 3} \right)}^2}}}\)
\(={{{x^3} + 9{x^2} + 27x + 27 + 6{x^2} - 54 - \left( {{x^3} - 9{x^2} + 27x - 27} \right)} \over {{{\left( {x - 3} \right)}^2}{{\left( {x + 3} \right)}^2}}}\)
\(={{24{x^2}} \over {{{\left( {x - 3} \right)}^2}{{\left( {x + 3} \right)}^2}}}\)
\(={{24{x^2}} \over {{{\left( {{x^2} - 9} \right)}^2}}}\)
+Ngoặc vuông thứ hai:
\(1:\left( {{{24{x^2}} \over {{x^4} - 81}} - {{12} \over {{x^2} + 9}}} \right) = 1:\left[ {{{24{x^2}} \over {\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)}} - {{12} \over {{x^2} + 9}}} \right]\)
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\(=1:\left( {{{24{x^2} - 12\left( {{x^2} - 9} \right)} \over {\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)}}} \right)\)
\(=1:{{12{x^2} + 108} \over {\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)}}\)
\(=1.{{\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)} \over {12{x^2} + 108}}\)
\(={{\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)} \over {12{x^2} + 108}}\)
\(={{\left( {{x^2} - 9} \right)\left( {{x^2} + 9} \right)} \over {12\left( {{x^2} + 9} \right)}}\)
\(={{{x^2} - 9} \over {12}}\)
Nên
\(\left[ {{{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {{x^2} - 9}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}} \right]\left[ {1:\left( {{{24{x^2}} \over {{x^4} - 81}} - {{12} \over {{x^2} + 9}}} \right)} \right]\)
\(=\left[ {{{x + 3} \over {{{\left( {x - 3} \right)}^2}}} + {6 \over {{x^2} - 9}} - {{x - 3} \over {{{\left( {x + 3} \right)}^2}}}} \right]{{24{x^2}} \over {{{\left( {{x^2} - 9} \right)}^2}}}.{{{x^2} - 9} \over {12}}\)
\(= {{2{x^2}} \over {{x^2} - 9}}\left[ {1:\left( {{{24{x^2}} \over {{x^4} - 81}} - {{12} \over {{x^2} + 9}}} \right)} \right]\)
Tại \(x = - {1 \over 3}\) giá trị của biểu thức là:
\({{2{{\left( { - {1 \over 3}} \right)}^2}} \over {{{\left( { - {1 \over 3}} \right)}^2} - 9}} = {{2.{1 \over 9}} \over {{1 \over 9} - 9}} = {{{2 \over 9}} \over { - {{80} \over 9}}} = - {1 \over {40}}\)