Câu 39 trang 34 SBT Toán 8 tập 1: Thực hiện phép chia phân thức :...

Thực hiện phép chia phân thức :

a. ({{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}:{{{x^2} – 4x + 4} over {{x^2} + 3x}})

b. ({{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}:{{{x^2} + 7x + 12} over {{x^2} – 9x + 14}})

Giải:

a. ({{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}:{{{x^2} – 4x + 4} over {{x^2} + 3x}})( = {{{x^2} – 5x + 6} over {{x^2} + 7x + 12}}.{{{x^2} + 3x} over {{x^2} – 4x + 4}})

( = {{left( {{x^2} – 5x + 6} ight).xleft( {x + 3} ight)} over {left( {{x^2} + 7x + 12} ight){{left( {x – 2} ight)}^2}}} = {{left( {{x^2} – 2x – 3x + 6} ight).xleft( {x + 3} ight)} over {left( {{x^2} + 3x + 4x + 12} ight){{left( {x – 2} ight)}^2}}})

( = {{left[ {xleft( {x – 2} ight) – 3left( {x – 2} ight)} ight].xleft( {x + 3} ight)} over {left[ {xleft( {x + 3} ight) + 4left( {x + 3} ight)} ight]{{left( {x – 2} ight)}^2}}})

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

b. ({{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}:{{{x^2} + 7x + 12} over {{x^2} – 9x + 14}})( = {{{x^2} + 2x – 3} over {{x^2} + 3x – 10}}.{{{x^2} – 9x + 14} over {{x^2} + 7x + 12}})

(eqalign{  &  = {{left( {{x^2} + 2x – 3} ight)left( {{x^2} – 9x + 14} ight)} over {left( {{x^2} + 3x – 10} ight)left( {{x^2} + 7x + 12} ight)}} = {{left( {{x^2} + 3x – x – 3} ight)left( {{x^2} – 7x – 2x + 14} ight)} over {left( {{x^2} + 5x – 2x + 10} ight)left( {{x^2} + 3x + 4x + 12} ight)}}  cr  &  = {{left[ {xleft( {x + 3} ight) – left( {x + 3} ight)} ight]left[ {xleft( {x – 7} ight) – 2left( {x – 7} ight)} ight]} over {left[ {xleft( {x + 5} ight) – 2left( {x + 5} ight)} ight]left[ {xleft( {x + 3} ight) + 4left( {x + 3} ight)} ight]}}  cr  &  = {{left( {x + 3} ight)left( {x – 1} ight)left( {x – 7} ight)left( {x – 2} ight)} over {left( {x + 5} ight)left( {x – 2} ight)left( {x + 3} ight)left( {x + 4} ight)}} = {{left( {x – 1} ight)left( {x – 7} ight)} over {left( {x + 5} ight)left( {x + 4} ight)}} cr} )