Bài 35 trang 50 sgk Toán 8 tập 1, Thực hiện các phép tính:...

Thực hiện các phép tính:

a)({{x + 1} over {x – 3}} – {{1 – x} over {x + 3}} – {{2xleft( {1 – x} ight)} over {9 – {x^2}}})

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

Hướng dẫn làm bài:

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

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

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

( = {{{x^2} + 4x + 3 + {x^2} – 4x + 3 + 2x – 2{x^2}} over {left( {x – 3} ight)left( {x + 3} ight)}})

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

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

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

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

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

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

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

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