Câu 24 trang 30 Sách bài tập Toán 8 tập 1: Làm tính nhân phân thức :...

Làm tính nhân phân thức :

a. ({{3x – 2} over {2xy}} – {{7x – 4} over {2xy}})

b. ({{3x + 5} over {4{x^3}y}} – {{5 – 15x} over {4{x^3}y}})

c. ({{4x + 7} over {2x + 2}} – {{3x + 6} over {2x + 2}})

d. ({{9x + 5} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}} – {{5x – 7} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}})

e. ({{xy} over {{x^2} – {y^2}}} – {{{x^2}} over {{y^2} – {x^2}}})

f. ({{5x + {y^2}} over {{x^2}y}} – {{5y – {x^2}} over {x{y^2}}})

g. ({x over {5x + 5}} – {x over {10x – 10}})

h. ({{x + 9} over {{x^2} – 9}} – {3 over {{x^2} + 3x}})

Giải:

a.  ({{3x – 2} over {2xy}} – {{7x – 4} over {2xy}})( = {{3x – 2} over {2xy}} + {{4 – 7x} over {2xy}} = {{3x – 2 + 4 – 7x} over {2xy}} = {{2left( {1 – 2x} ight)} over {2xy}} = {{1 – 2x} over {xy}})

b. ({{3x + 5} over {4{x^3}y}} – {{5 – 15x} over {4{x^3}y}})( = {{3x + 5} over {4{x^3}y}} + {{15x – 5} over {4{x^3}y}} = {{3x + 5 + 15x – 5} over {4{x^3}y}} = {{18x} over {4{x^3}y}} = {9 over {2{x^2}y}})

c. ({{4x + 7} over {2x + 2}} – {{3x + 6} over {2x + 2}})( = {{4x + 7} over {2x + 2}} + {{ – left( {3x + 6} ight)} over {2x + 2}} = {{4x + 7 – 3x – 6} over {2x + 2}} = {{x + 1} over {2left( {x + 1} ight)}} = {1 over 2})

d. ({{9x + 5} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}} – {{5x – 7} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}})( = {{9x + 5} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}} + {{7 – 5x} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}})

( = {{9x + 5 + 7 – 5x} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}} = {{4left( {x + 3} ight)} over {2left( {x – 1} ight){{left( {x + 3} ight)}^2}}} = {2 over {left( {x – 1} ight)left( {x + 3} ight)}})

e. ({{xy} over {{x^2} – {y^2}}} – {{{x^2}} over {{y^2} – {x^2}}})( = {{xy} over {{x^2} – {y^2}}} + {{{x^2}} over {{x^2} – {y^2}}} = {{xy + {x^2}} over {{x^2} – {y^2}}} = {{xleft( {x + y} ight)} over {left( {x + y} ight)left( {x – y} ight)}} = {x over {x – y}})

f. ({{5x + {y^2}} over {{x^2}y}} – {{5y – {x^2}} over {x{y^2}}})( = {{5x + {y^2}} over {{x^2}y}} + {{{x^2} – 5y} over {x{y^2}}} = {{yleft( {5x + {y^2}} ight)} over {{x^2}{y^2}}} + {{xleft( {{x^2} – 5y} ight)} over {{x^2}{y^2}}})

( = {{5xy + {y^3} + {x^3} – 5xy} over {{x^2}{y^2}}} = {{{x^3} + {y^3}} over {{x^2}{y^2}}})

g. ({x over {5x + 5}} – {x over {10x – 10}})( = {x over {5left( {x + 1} ight)}} + {{ – x} over {10left( {x – 1} ight)}} = {{2xleft( {x – 1} ight)} over {10left( {x + 1} ight)left( {x – 1} ight)}} + {{ – xleft( {x + 1} ight)} over {10left( {x + 1} ight)left( {x – 1} ight)}})

( = {{2{x^2} – 2x – {x^2} – x} over {10left( {x + 1} ight)left( {x – 1} ight)}} = {{{x^2} – 3x} over {10left( {x + 1} ight)left( {x – 1} ight)}})

h. ({{x + 9} over {{x^2} – 9}} – {3 over {{x^2} + 3x}})( = {{x + 9} over {left( {x + 3} ight)left( {x – 3} ight)}} + {{ – 3} over {xleft( {x + 3} ight)}} = {{xleft( {x + 9} ight)} over {xleft( {x + 3} ight)left( {x – 3} ight)}} + {{ – 3left( {x – 3} ight)} over {xleft( {x + 3} ight)left( {x – 3} ight)}})

( = {{{x^2} + 9x – 3x + 9} over {xleft( {x + 3} ight)left( {x – 3} ight)}} = {{{x^2} + 6x + 9} over {xleft( {x + 3} ight)left( {x – 3} ight)}} = {{{{left( {x + 3} ight)}^2}} over {xleft( {x + 3} ight)left( {x – 3} ight)}} = {{x + 3} over {xleft( {x – 3} ight)}})