Chứng minh các đẳng thức sau:
a) \(\left( {{a^2} - {1 \over a}} \right).\left( {{{a + 1} \over {{a^2} + 1 + a}} - {1 \over {1 - a}}} \right) = 2a + 1\) ;
b) \({3 \over {{x^2} - 3x}} - {{{x^2}} \over {3 - x}} = x + 3 + {{9x + 3} \over {{x^2} - 3x}}\) ;
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\(\eqalign{ & a)\,\,\left( {{a^2} - {1 \over a}} \right)\left( {{{a + 1} \over {{a^2} + 1 + a}} - {1 \over {1 - a}}} \right) = 2a + 1 \cr & VT = {{{a^3} - 1} \over a}.{{\left( {a + 1} \right)\left( {1 - a} \right) - \left( {{a^2} + 1 + a} \right)} \over {\left( {{a^2} + 1 + a} \right)\left( {1 - a} \right)}} \cr & \,\,\,\,\,\,\, = {{ - \left( {1 - {a^3}} \right)} \over a}.{{1 - {a^2} - {a^2} - 1 - a} \over {\left( {{a^2} + 1 + a} \right)\left( {1 - a} \right)}} \cr & \,\,\,\,\,\,\, = {{ - \left( { - 2{a^2} - a} \right)} \over a} = {{a\left( {2a + 1} \right)} \over a} = 2a + 1 = VP \cr & b)\,\,{3 \over {{x^2} - 3x}} - {{{x^2}} \over {3 - x}} = x + 3 + {{9x + 3} \over {{x^2} - 3x}} \cr & VT = {3 \over {x\left( {x - 3} \right)}} + {{{x^2}} \over {x - 3}} = {{3 + {x^3}} \over {x\left( {x + 3} \right)}} \cr & VP = {{\left( {x + 3} \right)\left( {{x^2} - 3x} \right) + 9x + 3} \over {{x^2} - 3x}} \cr & \,\,\,\,\,\,\, = {{{x^3} - 3{x^2} + 3{x^2} - 9x + 9x + 3} \over {{x^2} - 3x}} = {{{x^3} + 3} \over {x\left( {x - 3} \right)}} = VT \cr} \)