Advertisements (Quảng cáo)
a) \({{x + y} \over {x – y}} + {{{x^2} – 4{y^2}} \over {{x^2} – {y^2}}} – {{x – 3y} \over {x + y}}\) ;
b) \({1 \over {2z – 3}} – {2 \over {3 – 2z}} + {{18} \over {9 – 4{z^2}}}\) ;
c) \({1 \over {{a^2} – 5a – 6}} – {a \over {a – 6}}\) ;
d) \({x \over {x + y}} + {4 \over {{x^2} + 3xy + 2{y^2}}} – {{3x} \over {x + 2y}}\) .
Advertisements (Quảng cáo)
\(\eqalign{ & a)\,\,{{x + y} \over {x – y}} + {{{x^2} – 4{y^2}} \over {{x^2} – {y^2}}} – {{x – 3y} \over {x + y}} \cr & = {{x + y} \over {x – y}} + {{{x^2} – 4{y^2}} \over {\left( {x – y} \right)\left( {x + y} \right)}} – {{x – 3y} \over {x + y}} \cr & = {{{{\left( {x + y} \right)}^2}} \over {\left( {x – y} \right)\left( {x + y} \right)}} + {{{x^2} – 4{y^2}} \over {\left( {x – y} \right)\left( {x + y} \right)}} – {{\left( {x – 3y} \right)\left( {x – y} \right)} \over {\left( {x – y} \right)\left( {x + y} \right)}} \cr & = {{{{\left( {x + y} \right)}^2} + {x^2} – 4{y^2} – \left( {x – 3y} \right)\left( {x – y} \right)} \over {\left( {x – y} \right)\left( {x + y} \right)}} \cr & = {{{x^2} + 2xy + {y^2} + {x^2} – 4{y^2} – {x^2} + xy + 3xy – 3{y^2}} \over {\left( {x – 3y} \right)\left( {x – y} \right)}} \cr & = {{{x^2} + 6xy – 6{y^2}} \over {\left( {x – 3y} \right)\left( {x – y} \right)}} \cr & b)\,\,{1 \over {2z – 3}} – {2 \over {3 – 2z}} + {{18} \over {9 – 4{z^2}}} \cr & = {1 \over {2z – 3}} – {2 \over {3 – 2z}} + {{18} \over {\left( {3 – 2z} \right)\left( {3 + 2z} \right)}} \cr & = {{ – 3} \over {3 – 2z}} + {{18} \over {\left( {3 – 2z} \right)\left( {3 + 2z} \right)}} = {{ – 3\left( {3 + 2z} \right) + 18} \over {\left( {3 – 2z} \right)\left( {3 + 2z} \right)}} \cr & = {{9 – 6z} \over {\left( {3 – 2z} \right)\left( {3 + 2z} \right)}} = {{3\left( {3 – 2z} \right)} \over {\left( {3 – 2z} \right)\left( {3 + 2z} \right)}} = {3 \over {3 + 2z}} \cr & c)\,\,{1 \over {{a^2} – 5a – 6}} – {a \over {a – 6}} \cr & = {1 \over {\left( {a + 1} \right)\left( {a – 6} \right)}} – {a \over {a – 6}} \cr & = {{1 – a\left( {a + 1} \right)} \over {\left( {a + 1} \right)\left( {a – 6} \right)}} = {{ – {a^2} – a + 1} \over {\left( {a + 1} \right)\left( {a – 6} \right)}} \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 {\left( {x + y} \right)\left( {x + 2y} \right)}} + {{ – 3x} \over {x + 2y}} \cr & = {{x\left( {x + 2y} \right) + 4 – 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)}} = {{ – 2{x^2} – xy + 4} \over {\left( {x + y} \right)\left( {x + 2y} \right)}} \cr} \)