Câu 18 trang 28 SBT Toán 8 tập 1: Cộng các phân thức khác mẫu thức:...

Cộng các phân thức khác mẫu thức:

a. ({5 over {6{x^2}y}} + {7 over {12x{y^2}}} + {{11} over {18xy}})

b. ({{4x + 2} over {15{x^3}y}} + {{5y – 3} over {9{x^2}y}} + {{x + 1} over {5x{y^3}}})

c. ({3 over {2x}} + {{3x – 3} over {2x – 1}} + {{2{x^2} + 1} over {4{x^2} – 2x}})

d. ({{{x^3} + 2x} over {{x^3} + 1}} + {{2x} over {{x^2} – x + 1}} + {1 over {x + 1}})

Giải:

a. ({5 over {6{x^2}y}} + {7 over {12x{y^2}}} + {{11} over {18xy}})( = {{30y} over {36{x^2}{y^2}}} + {{21x} over {36{x^2}{y^2}}} + {{22xy} over {36{x^2}{y^2}}} = {{30y + 21x + 22xy} over {36{x^2}{y^2}}})

b. ({{4x + 2} over {15{x^3}y}} + {{5y – 3} over {9{x^2}y}} + {{x + 1} over {5x{y^3}}})(eqalign{  &  = {{3{y^2}left( {4x + 2} ight)} over {45{x^3}{y^3}}} + {{5x{y^2}left( {5y – 3} ight)} over {45{x^3}{y^3}}} + {{9{x^2}left( {x + 1} ight)} over {45{x^3}{y^3}}}  cr  &  = {{12x{y^2} + 6{y^2} + 25x{y^3} – 15x{y^2} + 9{x^3} + 9{x^2}} over {45{x^3}{y^3}}} = {{6{y^2} + 25x{y^3} – 3x{y^2} + 9{x^3} + 9{x^2}} over {45{x^3}{y^3}}} cr} )

c. ({3 over {2x}} + {{3x – 3} over {2x – 1}} + {{2{x^2} + 1} over {4{x^2} – 2x}})( = {3 over {2x}} + {{3x – 3} over {2x – 1}} + {{2{x^2} + 1} over {2xleft( {2x – 1} ight)}})

(eqalign{  &  = {{3left( {2x – 1} ight)} over {2xleft( {2x – 1} ight)}} + {{2xleft( {3x – 3} ight)} over {2xleft( {2x – 1} ight)}} + {{2{x^2} + 1} over {2xleft( {2x – 1} ight)}} = {{6x – 3 + 6{x^2} – 6x + 2{x^2} + 1} over {2xleft( {2x – 1} ight)}}  cr  &  = {{8{x^2} – 2} over {2xleft( {2x – 1} ight)}} = {{2left( {4{x^2} – 1} ight)} over {2xleft( {2x – 1} ight)}} = {{left( {2x + 1} ight)left( {2x – 1} ight)} over {xleft( {2x – 1} ight)}} = {{2x + 1} over x} cr} )

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

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