Câu 45 trang 36 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} )