Câu 45 trang 36 Sách bài tập (SBT) Toán 8 tập 1

Thực hiện các phép tính sau :

a. \(\left( {{{5x + y} \over {{x^2} - 5xy}} + {{5x - y} \over {{x^2} + 5xy}}} \right).{{{x^2} - 25{y^2}} \over {{x^2} + {y^2}}}\)

b. \({{4xy} \over {{y^2} - {x^2}}}:\left( {{1 \over {{x^2} + 2xy + {y^2}}} - {1 \over {{x^2} - {y^2}}}} \right)\)

c. \(\left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {4{x^2} - {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)

d. \(\left( {{2 \over {x + 2}} - {4 \over {{x^2} + 4x + 4}}} \right):\left( {{2 \over {{x^2} - 4}} + {1 \over {2 - x}}} \right)\)

Giải:

a. \(\left( {{{5x + y} \over {{x^2} - 5xy}} + {{5x - y} \over {{x^2} + 5xy}}} \right).{{{x^2} - 25{y^2}} \over {{x^2} + {y^2}}}\)

\(\eqalign{  &  = \left[ {{{5x + y} \over {x\left( {x - 5y} \right)}} + {{5x - y} \over {x\left( {x + 5y} \right)}}} \right].{{{x^2} - 25{y^2}} \over {{x^2} + {y^2}}}  \cr  &  = {{\left( {5x + y} \right)\left( {x + 5y} \right) + \left( {5x - y} \right)\left( {x - 5y} \right)} \over {x\left( {x - 5y} \right)\left( {x + 5y} \right)}}.{{\left( {x - 5y} \right)\left( {x + 5y} \right)} \over {{x^2} + {y^2}}}  \cr  &  = {{5{x^2} + 25xy + xy + 5{y^2} + 5{x^2} - 25xy - xy + 5{y^2}} \over {x\left( {{x^2} + {y^2}} \right)}}  \cr  &  = {{10{x^2} + 10{y^2}} \over {x\left( {{x^2} + {y^2}} \right)}} = {{10\left( {{x^2} + {y^2}} \right)} \over {x\left( {{x^2} + {y^2}} \right)}} = {{10} \over x} \cr} \)

b. \({{4xy} \over {{y^2} - {x^2}}}:\left( {{1 \over {{x^2} + 2xy + {y^2}}} - {1 \over {{x^2} - {y^2}}}} \right)\)

\(\eqalign{  &  = {{4xy} \over {{y^2} - {x^2}}}:\left[ {{1 \over {{{\left( {x + y} \right)}^2}}} - {1 \over {\left( {x + y} \right)\left( {x - y} \right)}}} \right]  \cr  &  = {{4xy} \over {{y^2} - {x^2}}}:{{x - y - \left( {x + y} \right)} \over {{{\left( {x + y} \right)}^2}\left( {x - y} \right)}} = {{4xy} \over {{y^2} - {x^2}}}:{{ - 2y} \over {{{\left( {x + y} \right)}^2}\left( {x - y} \right)}} = {{4xy} \over {{y^2} - {x^2}}}.{{{{\left( {x + y} \right)}^2}\left( {y - x} \right)} \over {2y}}  \cr  &  = {{4xy{{\left( {x + y} \right)}^2}\left( {y - x} \right)} \over {\left( {y + x} \right)\left( {y - x} \right).2y}} = 2x\left( {x + y} \right) \cr} \)

c. \(\left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {4{x^2} - {y^2}}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{4{x^2} + 4xy + {y^2}} \over {16x}}\)

\(\eqalign{  &  = \left[ {{1 \over {{{\left( {2x - y} \right)}^2}}} + {2 \over {\left( {2x + y} \right)\left( {2x - y} \right)}} + {1 \over {{{\left( {2x + y} \right)}^2}}}} \right].{{{{\left( {2x + y} \right)}^2}} \over {16x}}  \cr  &  = {{{{\left( {2x + y} \right)}^2} + 2\left( {2x + y} \right)\left( {2x - y} \right) + {{\left( {2x - y} \right)}^2}} \over {{{\left( {2x + y} \right)}^2}.{{\left( {2x - y} \right)}^2}}}.{{{{\left( {2x + y} \right)}^2}} \over {16x}}  \cr  &  = {{{{\left[ {\left( {2x + y} \right) + \left( {2x - y} \right)} \right]}^2}} \over {16x{{\left( {2x - y} \right)}^2}}} = {{{{\left( {4x} \right)}^2}} \over {16x{{\left( {2x - y} \right)}^2}}} = {{16{x^2}} \over {16x{{\left( {2x - y} \right)}^2}}} = {x \over {{{\left( {2x - y} \right)}^2}}} \cr} \)

d. \(\left( {{2 \over {x + 2}} - {4 \over {{x^2} + 4x + 4}}} \right):\left( {{2 \over {{x^2} - 4}} + {1 \over {2 - x}}} \right)\)

\(\eqalign{  &  = \left[ {{2 \over {x + 2}} - {4 \over {{{\left( {x + 2} \right)}^2}}}} \right]:\left[ {{2 \over {\left( {x + 2} \right)\left( {x - 2} \right)}} - {1 \over {x - 2}}} \right]  \cr  &  = {{2\left( {x + 2} \right) - 4} \over {{{\left( {x + 2} \right)}^2}}}:{{2 - \left( {x + 2} \right)} \over {\left( {x + 2} \right)\left( {x - 2} \right)}} = {{2x + 4 - 4} \over {{{\left( {x + 2} \right)}^2}}}:{{2 - x - 2} \over {\left( {x + 2} \right)\left( {x - 2} \right)}}  \cr  &  = {{2x} \over {{{\left( {x + 2} \right)}^2}}}.{{\left( {x + 2} \right)\left( {x - 2} \right)} \over { - x}} = {{2\left( {x - 2} \right)} \over { - \left( {x + 2} \right)}} = {{2\left( {2 - x} \right)} \over {x + 2}} \cr} \)