Đề bài
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) \({x \over {(x - 2y)(z - x)}} + {{2y} \over {(2y - z)(x - 2y)}} + {z \over {(z - x)(2y - z)}}\) .
Lời giải chi tiết
\(\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} \)
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