Câu 14 trang 27 Sách bài tập (SBT) Toán 8 tập 1

Quy đồng mẫu thức các phân thức:

Quy đồng mẫu thức các phân thức:

a. ({{7x - 1} over {2{x^2} + 6x}},{{5 - 3x} over {{x^2} - 9}})

b. ({{x + 1} over {x - {x^2}}},{{x + 2} over {2 - 4x + 2{x^2}}})

c. ({{4{x^2} - 3x + 5} over {{x^3} - 1}},{{2x} over {{x^2} + x + 1}},{6 over {x - 1}})

d. ({7 over {5x}},{4 over {x - 2y}},{{x - y} over {8{y^2} - 2{x^2}}})

e. ({{5{x^2}} over {{x^3} + 6{x^2} + 12x + 8}},{{4x} over {{x^2} + 4x + 4}},{3 over {2x + 4}})

Giải:

a. (2{x^2} + 6x = 2xleft( {x + 3} ight);{x^2} - 9 = left( {x + 3} ight)left( {x - 3} ight))  MTC = (2xleft( {x + 3} ight)left( {x - 3} ight))

(eqalign{  & {{7x - 1} over {2{x^2} + 6x}} = {{7x - 1} over {2xleft( {x + 3} ight)}} = {{left( {7x - 1} ight)left( {x - 3} ight)} over {2xleft( {x + 3} ight)left( {x - 3} ight)}}  cr  & {{5 - 3x} over {{x^2} - 9}} = {{5 - 3x} over {left( {x + 3} ight)left( {x - 3} ight)}} = {{2xleft( {5 - 3x} ight)} over {2xleft( {x + 3} ight)left( {x - 3} ight)}} cr} )

b. (x - {x^2} = xleft( {1 - x} ight)); (2 - 4x + 2{x^2} = 2left( {1 - 2x + {x^2}} ight) = 2{left( {1 - x} ight)^2})

MTC = (2x{left( {1 - x} ight)^2})

(eqalign{  & {{x + 1} over {x - {x^2}}} = {{x + 1} over {xleft( {1 - x} ight)}} = {{left( {x + 1} ight).2left( {1 - x} ight)} over {xleft( {1 - x} ight).2left( {1 - x} ight)}} = {{2{{left( {1 - x} ight)}^2}} over {2x{{left( {1 - x} ight)}^2}}}  cr  & {{x + 2} over {2 - 4x + 2{x^2}}} = {{x + 2} over {2{{left( {1 - x} ight)}^2}}} = {{left( {x + 2} ight).x} over {2x{{left( {1 - x} ight)}^2}}} cr} )

c. ({x^3} - 1 = left( {x - 1} ight)left( {{x^2} + x + 1} ight)) MTC = ({x^3} - 1) ({{4{x^2} - 3x + 5} over {{x^3} - 1}});

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

d. (8{y^2} - 2{x^2} = 2left( {4{y^2} - {x^2}} ight) = 2left( {2y + x} ight)left( {2y - x} ight))

MTC = (10xleft( {2y + x} ight)left( {2y - x} ight))

(eqalign{  & {7 over {5x}} = {{7.2left( {2y + x} ight)left( {2y - x} ight)} over {5x.2left( {2y + x} ight)left( {2y - x} ight)}} = {{14left( {2y + x} ight)left( {2y - x} ight)} over {10xleft( {2y + x} ight)left( {2y - x} ight)}}  cr  & {4 over {x - 2y}} = {{ - 4} over {2y - x}} = {{ - 4.10xleft( {2y + x} ight)} over {left( {2y - x} ight).10xleft( {2y + x} ight)}} = {{ - 40xleft( {2y + x} ight)} over {10xleft( {2y + x} ight)left( {2y - x} ight)}}  cr  & {{x - y} over {8{y^2} - 2{x^2}}} = {{x - y} over {2left( {2y + x} ight)left( {2y - x} ight)}} = {{left( {x - y} ight).5x} over {2left( {2y + x} ight)left( {2y - x} ight).5x}}  cr  &  = {{5xleft( {x - y} ight)} over {10xleft( {2y + x} ight)left( {2y - x} ight)}} cr} )

e. (eqalign{  & {x^3} + 6{x^2} + 12x + 8 = {x^3} + 3{x^2}.2 + 3.x{.2^2} + {2^3} = {left( {x + 2} ight)^3}  cr  & {x^2} + 4x + 4 = {left( {x + 2} ight)^2};2x + 4 = 2left( {x + 2} ight) cr} )

MTC =(2{left( {x + 2} ight)^3})

(eqalign{  & {{5{x^2}} over {{x^3} + 6{x^2} + 12x + 8}} = {{5{x^2}} over {{{left( {x + 2} ight)}^3}}} = {{5{x^2}.2} over {{{left( {x + 2} ight)}^3}.2}} = {{10{x^2}} over {2{{left( {x + 2} ight)}^3}}}  cr  & {{4x} over {{x^2} + 4x + 4}} = {{4x} over {{{left( {x + 2} ight)}^2}}} = {{4x.2left( {x + 2} ight)} over {{{left( {x + 2} ight)}^2}.2left( {x + 2} ight)}} = {{8xleft( {x + 2} ight)} over {2{{left( {x + 2} ight)}^3}}}  cr  & {3 over {2x + 4}} = {3 over {2left( {x + 2} ight)}} = {{3{{left( {x + 2} ight)}^2}} over {2left( {x + 2} ight){{left( {x + 2} ight)}^2}}} = {{3{{left( {x + 2} ight)}^2}} over {2{{left( {x + 2} ight)}^3}}} cr} )