Rút gọn:
1. \(A = {{x + 8} \over {x – 2}}.{{3x – 7} \over {x + 10}} + {{x + 8} \over {x – 2}}.{{5 – 2x} \over {x + 10}}\)
2. \(B = \left( {{x \over {x + y}} + {x \over {x – y}}} \right).\left( {{x^2} – {y^2}} \right)\)
3. \(C = {{{x^2} + 3x} \over {{x^2} + 3x + 9}}.{{{x^3} – 27} \over {4x + 12}}.\)
1. \(A = {{x + 8} \over {x – 2}}\left( {{{3x – 7} \over {x + 10}} + {{5 – 2x} \over {x + 10}}} \right) \)
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\(\;\;\;\;\;\;\;\;= {{x + 8} \over {x – 2}}.{{3x – 7 + 5 – 2x} \over {x + 10}}\)
\( \;\;\;\;\;\;\;\;= {{x + 8} \over {x – 2}}.{{x – 2} \over {x + 10}} = {{x + 8} \over {x + 10}}.\)
2. \(B = {{x\left( {{x^2} – {y^2}} \right)} \over {x + y}} + {{y\left( {{x^2} – {y^2}} \right)} \over {x – y}}\)
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\(\;\;\;\;\;\;\;= x\left( {x – y} \right) + y\left( {x + y} \right) = {x^2} + {y^2}.\)
3. \(C = {{x\left( {x + 3} \right)\left( {x – 3} \right)\left( {{x^2} + 3x + 9} \right)} \over {\left( {{x^2} + 3x + 9} \right)4\left( {x + 3} \right)}} = {{x\left( {x – 3} \right)} \over 4} \)\(\;= {{{x^2} – 3x} \over 4}.\)