Đề bài - bài tập 10 trang 79 tài liệu dạy – học toán 8 tập 1
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\(\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} \) Đề bài 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) \({x \over {x + y}} + {4 \over {{x^2} + 3xy + 2{y^2}}} - {{3x} \over {x + 2y}}\) . Lời giải chi tiết \(\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} \)
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