Bài 58 trang 62 sgk Toán 8 tập 1, Thực hiện các phép tính sau:...

Thực hiện các phép tính sau:

a) (left( {{{2x + 1} over {2x – 1}} – {{2x – 1} over {2x + 1}}} ight):{{4x} over {10x – 5}})                     

b) (left( {{1 over {{x^2} + x}} – {{2 – x} over {x + 1}}} ight):left( {{1 over x} + x – 2} ight);)

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

Hướng dẫn làm bài:

a) (left( {{{2x + 1} over {2x – 1}} – {{2x – 1} over {2x + 1}}} ight):{{4x} over {10x – 5}} = {{{{left( {2x + 1} ight)}^2} – {{left( {2x – 1} ight)}^2}} over {left( {2x – 1} ight)left( {2x + 1} ight)}}.{{10x + 5} over {4x}})

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

=({{8x.5left( {2x + 1} ight)} over {left( {2x – 1} ight)left( {2x + 1} ight).4x}} = {{10} over {2x – 1}})

b) (left( {{1 over {{x^2} + x}} – {{2 – x} over {x + 1}}} ight):left( {{1 over x} + x – 2} ight))

=(left( {{1 over {xleft( {x + 1} ight)}} + {{x – 2} over {x + 1}}} ight):{{1 + {x^2} – 2x} over x})

=({{1 + xleft( {x – 2} ight)} over {xleft( {x + 1} ight)}}.{x over {{x^2} – 2x + 1}})

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

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

=({1 over {x – 1}} – {{{x^3} – x} over {{x^2} + 1}}.left[ {{1 over {{{left( {x – 1} ight)}^2}}} – {1 over {left( {x – 1} ight)left( {x + 1} ight)}}} ight])

=({1 over {x – 1}} – {{xleft( {{x^2} – 1} ight)} over {{x^2} + 1}}.{{x + 1 – left( {x – 1} ight)} over {{{left( {x – 1} ight)}^2}.left( {x + 1} ight)}})

=({1 over {x – 1}} – {{xleft( {x – 1} ight)left( {x + 1} ight)} over {{x^2} + 1}}.{{x + 1 – x + 1} over {{{left( {x – 1} ight)}^2}left( {x + 1} ight)}})

=({1 over {x – 1}} – {{xleft( {x – 1} ight)left( {x + 1} ight).2} over {left( {{x^2} + 1} ight){{left( {x – 1} ight)}^2}left( {x + 1} ight)}} = {1 over {x – 1}} – {{2x} over {left( {{x^2} + 1} ight)left( {x – 1} ight)}})

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