Câu 58 trang 39 Sách bài tập Toán 8 tập 1: Thực hiện các phép tính :...

Thực hiện các phép tính :

a. (left( {{9 over {{x^3} – 9x}} + {1 over {x + 3}}} ight):left( {{{x – 3} over {{x^2} + 3x}} – {x over {3x + 9}}} ight))

b. (left( {{2 over {x – 2}} – {2 over {x + 2}}} ight).{{{x^2} + 4x + 4} over 8})

c. (left( {{{3x} over {1 – 3x}} + {{2x} over {3x + 1}}} ight):{{6{x^2} + 10x} over {1 – 6x + 9{x^2}}})

d. (left( {{x over {{x^2} – 25}} – {{x – 5} over {{x^2} + 5x}}} ight):{{2x – 5} over {{x^2} + 5x}} + {x over {5 – x}})

e. (left( {{{{x^2} + xy} over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y over {{x^2} + {y^2}}}} ight):left( {{1 over {x – y}} – {{2xy} over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} ight))

Giải:

a. (left( {{9 over {{x^3} – 9x}} + {1 over {x + 3}}} ight):left( {{{x – 3} over {{x^2} + 3x}} – {x over {3x + 9}}} ight))

(eqalign{  &  = left[ {{9 over {xleft( {x + 3} ight)left( {x – 3} ight)}} + {1 over {x + 3}}} ight]:left[ {{{x – 3} over {xleft( {x + 3} ight)}} – {x over {3left( {x + 3} ight)}}} ight]  cr  &  = {{9 + xleft( {x – 3} ight)} over {xleft( {x + 3} ight)left( {x – 3} ight)}}:{{3left( {x – 3} ight) – {x^2}} over {3xleft( {x + 3} ight)}} = {{{x^2} – 3x + 9} over {xleft( {x + 3} ight)left( {x – 3} ight)}}.{{3xleft( {x + 3} ight)} over {3x – 9 – {x^2}}}  cr  &  = {{3left( {{x^2} – 3x + 9} ight)} over {left( {3 – x} ight)left( {{x^2} – 3x + 9} ight)}} = {3 over {3 – x}} cr} )

b. (left( {{2 over {x – 2}} – {2 over {x + 2}}} ight).{{{x^2} + 4x + 4} over 8})( = {{2left( {x + 2} ight) – 2left( {x – 2} ight)} over {left( {x – 2} ight)left( {x + 2} ight)}}.{{{{left( {x + 2} ight)}^2}} over 8})

( = {{2x + 4 – 2x + 4} over {left( {x – 2} ight)left( {x + 2} ight)}}.{{{{left( {x + 2} ight)}^2}} over 8} = {8 over {left( {x – 2} ight)left( {x + 2} ight)}}.{{{{left( {x + 2} ight)}^2}} over 8} = {{x + 2} over {x – 2}})

c. (left( {{{3x} over {1 – 3x}} + {{2x} over {3x + 1}}} ight):{{6{x^2} + 10x} over {1 – 6x + 9{x^2}}})( = {{3xleft( {3x + 1} ight) + 2xleft( {1 – 3x} ight)} over {left( {1 – 3x} ight)left( {1 + 3x} ight)}}:{{2xleft( {3x + 5} ight)} over {{{left( {1 – 3x} ight)}^2}}})

(eqalign{  &  = {{9{x^2} + 3x + 2x – 6{x^2}} over {left( {1 – 3x} ight)left( {1 + 3x} ight)}}.{{{{left( {1 – 3x} ight)}^2}} over {2xleft( {3x + 5} ight)}} = {{xleft( {3x + 5} ight)} over {left( {1 – 3x} ight)left( {1 + 3x} ight)}}.{{{{left( {1 – 3x} ight)}^2}} over {2xleft( {3x + 5} ight)}}  cr  &  = {{1 – 3x} over {2left( {1 + 3x} ight)}} cr} )

d. (left( {{x over {{x^2} – 25}} – {{x – 5} over {{x^2} + 5x}}} ight):{{2x – 5} over {{x^2} + 5x}} + {x over {5 – x}})

(eqalign{ &  = left[ {{x over {left( {x + 5} ight)left( {x – 5} ight)}} – {{x – 5} over {xleft( {x + 5} ight)}}} ight]:{{2x – 5} over {xleft( {x + 5} ight)}} + {x over {5 – x}}  cr  &  = {{{x^2} – {{left( {x – 5} ight)}^2}} over {xleft( {x + 5} ight)left( {x – 5} ight)}}.{{xleft( {x + 5} ight)} over {2x – 5}} + {x over {5 – x}}  cr  &  = {{{x^2} – {x^2} + 10x – 25} over {left( {x – 5} ight)left( {2x – 5} ight)}} + {x over {5 – x}} = {{5left( {2x – 5} ight)} over {left( {x – 5} ight)left( {2x – 5} ight)}} – {x over {x – 5}}  cr  &  = {5 over {x – 5}} – {x over {x – 5}} = {{5 – x} over {x – 5}} = {{ – left( {x – 5} ight)} over {x – 5}} =  – 1 cr} )

e. (left( {{{{x^2} + xy} over {{x^3} + {x^2}y + x{y^2} + {y^3}}} + {y over {{x^2} + {y^2}}}} ight):left( {{1 over {x – y}} – {{2xy} over {{x^3} – {x^2}y + x{y^2} – {y^3}}}} ight))

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