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

Thực hiện các phép tính sau :

Thực hiện các phép tính sau :

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

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

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

Giải:

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

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

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

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

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

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

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