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Thực hiện phép tính và rút gọn:
a) \({{x + 1} \over {2x + 6}} + {{2x + 3} \over {x(x + 3)}}\) ;
b) \({{1 – 3x} \over {2x}} + {{3x – 2} \over {2x – 1}} + {{3x – 2} \over {2x – 4{x^2}}}\) ;
c) \({x \over {(x – 2y)(z – x)}} + {{2y} \over {(2y – z)(x – 2y)}} + {z \over {(z – x)(2y – z)}}\) .
\(\eqalign{ & a)\,\,{{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}} \cr & \,\,\,\,\, = {{x + 1} \over {2\left( {x + 3} \right)}} + {{2x + 3} \over {x\left( {x + 3} \right)}} \cr & \,\,\,\,\, = {{\left( {x + 1} \right)x} \over {2\left( {x + 3} \right)x}} + {{2\left( {2x + 3} \right)} \over {2\left( {x + 3} \right)x}} \cr & \,\,\,\,\, = {{{x^2} + x + 4x + 6} \over {2\left( {x + 3} \right)x}} = {{{x^2} + 5x + 6} \over {2\left( {x + 3} \right)x}} \cr & \,\,\,\,\, = {{\left( {x + 3} \right)\left( {x + 2} \right)} \over {2\left( {x + 3} \right)x}} = {{x + 2} \over {2x}} \cr & b)\,\,{{1 – 3x} \over {2x}} + {{3x – 2} \over {2x – 1}} + {{3x – 2} \over {2x – 4{x^2}}} \cr & \,\,\,\,\, = {{\left( {1 – 3x} \right)} \over {2x}} + {{ – \left( {3x – 2} \right)} \over {1 – 2x}} + {{3x – 2} \over {2x\left( {1 – 2x} \right)}} \cr & \,\,\,\,\, = {{\left( {1 – 3x} \right)\left( {1 – 2x} \right)} \over {2x\left( {1 – 2x} \right)}} + {{ – \left( {3x – 2} \right).2x} \over {2x\left( {1 – 2x} \right)}} + {{3x – 2} \over {2x\left( {1 – 2x} \right)}} \cr & \,\,\,\,\, = {{1 – 2x – 3x + 6{x^2} – 6{x^2} + 4x + 3x – 2} \over {2x\left( {1 – 2x} \right)}} \cr & \,\,\,\,\, = {{2x – 1} \over {2x\left( {1 – 2x} \right)}} = {{ – \left( {1 – 2x} \right)} \over {2x\left( {1 – 2x} \right)}} = {{ – 1} \over {2x}} \cr & c)\,\,{x \over {\left( {x – 2y} \right)\left( {z – x} \right)}} + {{2y} \over {\left( {2y – z} \right)\left( {x – 2y} \right)}} + {z \over {\left( {z – x} \right)\left( {2y – z} \right)}} \cr & \,\,\,\,\, = {{x\left( {2y – z} \right)} \over {\left( {x – 2y} \right)\left( {z – x} \right)\left( {2y – z} \right)}} + {{2y\left( {z – x} \right)} \over {\left( {x – 2y} \right)\left( {z – x} \right)\left( {2y – z} \right)}} + {{z\left( {x – 2y} \right)} \over {\left( {x – 2y} \right)\left( {z – x} \right)\left( {2y – z} \right)}} \cr & \,\,\,\,\, = {{2xy – xz + 2yz – 2xy + zx – 2yz} \over {\left( {x – 2y} \right)\left( {z – x} \right)\left( {2y – z} \right)}} \cr & \,\,\,\,\, = {0 \over {\left( {x – 2y} \right)\left( {z – x} \right)\left( {2y – z} \right)}} = 0 \cr} \)