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Thực hiện phép tính và rút gọn:
a) \({2 \over {a + 3}} + {3 \over {(a + 3)(a + 1)}}\) ;
b) \({1 \over {{p^2} – 4p – 5}} + {{2p} \over {p – 5}}\) ;
c) \(x + {{3x} \over {x + 2}} + 2\) ;
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d) \({x \over {x + y}} + {4 \over {{x^2} + 3xy + 2{y^2}}} + {{ – 3x} \over {x + 2y}}\) .
\(\eqalign{ & a)\,\,{2 \over {a + 3}} + {3 \over {\left( {a + 3} \right)\left( {a + 1} \right)}} \cr & \,\,\,\,\, = {{2\left( {a + 1} \right)} \over {\left( {a + 3} \right)\left( {a + 1} \right)}} + {3 \over {\left( {a + 3} \right)\left( {a + 1} \right)}} \cr & \,\,\,\,\, = {{2a + 5} \over {\left( {a + 3} \right)\left( {a + 1} \right)}} \cr & b)\,\,{1 \over {{p^2} – 4p – 5}} + {{2p} \over {p – 5}} \cr & \,\,\,\,\, = {1 \over {{p^2} + p – 5p – 5}} + {{2p} \over {p – 5}} \cr & \,\,\,\, = {1 \over {p\left( {p + 1} \right) – 5\left( {p + 1} \right)}} + {{2p} \over {p – 5}} \cr & \,\,\,\,\, = {1 \over {\left( {p + 1} \right)\left( {p – 5} \right)}} + {{2p\left( {p + 1} \right)} \over {\left( {p – 5} \right)\left( {p + 1} \right)}} \cr & \,\,\,\,\, = {{1 + 2{p^2} + 2p} \over {\left( {p + 1} \right)\left( {p – 5} \right)}} \cr & c)\,\,x + {{3x} \over {x + 2}} + 2 = {{x\left( {x + 2} \right)} \over {x + 2}} + {{3x} \over {x + 2}} + {{2\left( {x + 2} \right)} \over {x + 2}} \cr & \,\,\,\,\, = {{{x^2} + 2x + 3x + 2x + 4} \over {x + 2}} = {{{x^2} + 7x + 4} \over {x + 2}} \cr & d)\,\,{x \over {x + y}} + {4 \over {{x^2} + 3xy + 2{y^2}}} + {{ – 3x} \over {x + 2y}} \cr & \,\,\,\,\, = {x \over {x + y}} + {4 \over {{x^2} + xy + 2xy + 2{y^2}}} + {{ – 3x} \over {x + 2y}} \cr & \,\,\,\, = {x \over {x + y}} + {4 \over {x\left( {x + y} \right) + 2y\left( {x + y} \right)}} + {{ – 3x} \over {x + 2y}} \cr & \,\,\,\,\, = {x \over {x + y}} + {4 \over {\left( {x + y} \right)\left( {x + 2y} \right)}} + {{ – 3x} \over {x + 2y}} \cr & \,\,\,\,\, = {{x\left( {x + 2y} \right)} \over {\left( {x + y} \right)\left( {x + 2y} \right)}} + {4 \over {\left( {x + y} \right)\left( {x + 2y} \right)}} + {{ – 3x\left( {x + y} \right)} \over {\left( {x + y} \right)\left( {x + 2y} \right)}} \cr & \,\,\,\,\, = {{{x^2} + 2xy + 4 – 3{x^2} – 3xy} \over {\left( {x + y} \right)\left( {x + 2y} \right)}} \cr & \,\,\,\,\, = {{ – 2{x^2} – xy + 4} \over {\left( {x + y} \right)\left( {x + 2y} \right)}} \cr} \)