Câu 13 trang 27 Sách bài tập (SBT) Toán 8 tập 1
Quy đồng mẫu thức các phân thức:
a. ({{25} over {14{x^2}y}},{{14} over {21x{y^5}}})
b. ({{11} over {102{x^4}y}},{3 over {34x{y^3}}})
c. ({{3x + 1} over {12x{y^4}}},{{y – 2} over {9{x^2}{y^3}}})
d. ({1 over {6{x^3}{y^2}}},{{x + 1} over {9{x^2}{y^4}}},{{x – 1} over {4x{y^3}}})
e. ({{3 + 2x} over {10{x^4}y}},{5 over {8{x^2}{y^2}}},{2 over {3x{y^5}}})
f. ({{4x – 4} over {2xleft( {x + 3} right)}},{{x – 3} over {3xleft( {x + 1} right)}})
g. ({{2x} over {{{left( {x + 2} right)}^3}}},{{x – 2} over {2x{{left( {x + 2} right)}^2}}})
h. ({5 over {3{x^3} – 12x}},{3 over {left( {2x + 4} right)left( {x + 3} right)}})
Giải:
a. MTC ( = 42{x^2}{y^5})
({{14} over {21x{y^5}}} = {2 over {3x{y^5}}})( = {{2.14x} over {3x{y^5}.14x}} = {{28x} over {42{x^2}{y^5}}}); ({{25} over {14{x^2}y}} = {{25.{3y^4}} over {14{x^2}y.{3y^4}}} = {{75{y^4}} over {42{x^2}{y^5}}})
b. MTC = (102{x^4}{y^3})
({{11} over {102{x^4}y}} = {{11.{y^2}} over {102{x^4}y.{y^2}}} = {{11{y^2}} over {102{x^4}{y^3}}}); ({3 over {34x{y^3}}} = {{3.3{x^3}} over {34x{y^3}.3{x^3}}} = {{9{x^3}} over {102{x^4}{y^3}}})
c. MTC = (36{x^2}{y^4})
({{3x + 1} over {12x{y^4}}} = {{left( {3x + 1} right).3x} over {12x{y^4}.3x}} = {{9{x^2} + 3x} over {36{x^2}{y^4}}}); ({{y – 2} over {9{x^2}{y^3}}} = {{left( {y – 2} right).4y} over {9{x^2}{y^3}.4y}} = {{4{y^2} – 8y} over {36{x^2}{y^4}}})
d. MTC = (36{x^3}{y^4})
({1 over {6{x^3}{y^2}}} = {{1.6{y^2}} over {6{x^3}{y^2}.6{y^2}}} = {{6{y^2}} over {36{x^3}{y^4}}}); ({{x + 1} over {9{x^2}{y^4}}} = {{left( {x + 1} right).4x} over {9{x^2}{y^4}.4x}} = {{4{x^2} + 4x} over {36{x^3}{y^4}}})
({{x – 1} over {4x{y^3}}} = {{left( {x – 1} right).9{x^2}y} over {4x{y^3}.9{x^2}y}} = {{9{x^3}y – 9{x^2}y} over {36{x^3}{y^4}}})
e. MTC = (120{x^4}{y^5})
({{3 + 2x} over {10{x^4}y}} = {{left( {3 + 2x} right).12{y^4}} over {10{x^4}y.12{y^4}}} = {{36{y^4} + 24x{y^4}} over {120{x^4}{y^5}}})
({5 over {8{x^2}{y^2}}} = {{5.15{x^2}{y^3}} over {8{x^2}{y^2}.15{x^2}{y^3}}} = {{75{x^2}{y^3}} over {120{x^4}{y^5}}})
({2 over {3x{y^5}}} = {{2.40{x^3}} over {3x{y^5}.40{x^3}}} = {{80{x^3}} over {120{x^4}{y^5}}})
f. MTC = (3xleft( {x + 3} right)left( {x + 1} right)) Vì ({{4x – 4} over {2xleft( {x + 3} right)}} = {{2left( {x – 1} right)} over {xleft( {x + 3} right)}})
({{4x – 4} over {2xleft( {x + 3} right)}} = {{2left( {x – 1} right)} over {xleft( {x + 3} right)}} = {{2left( {x – 1} right).3left( {x + 1} right)} over {xleft( {x + 3} right).3left( {x + 1} right)}} = {{6left( {{x^2} – 1} right)} over {3xleft( {x + 3} right)left( {x + 1} right)}})
({{x – 3} over {3xleft( {x + 1} right)}} = {{left( {x – 3} right)left( {x + 3} right)} over {3xleft( {x + 1} right)left( {x + 3} right)}} = {{{x^2} – 9} over {3xleft( {x + 1} right)left( {x + 3} right)}})
g. MTC = (2x{left( {x + 2} right)^3})
({{2x} over {{{left( {x + 2} right)}^3}}} = {{2x.2x} over {2x{{left( {x + 2} right)}^3}}} = {{4{x^2}} over {2x{{left( {x + 2} right)}^3}}})
({{x – 2} over {2x{{left( {x + 2} right)}^2}}} = {{left( {x – 2} right)left( {x + 2} right)} over {2x{{left( {x + 2} right)}^2}left( {x + 2} right)}} = {{{x^2} – 4} over {2x{{left( {x + 2} right)}^3}}})
h. (3{x^3} – 12x = 3xleft( {{x^2} – 4} right) = 3xleft( {x – 2} right)left( {x + 2} right))
(left( {2x + 4} right)left( {x + 3} right) = 2left( {x + 2} right)left( {x + 3} right))
MTC = (6xleft( {x – 2} right)left( {x + 2} right)left( {x + 3} right))
(eqalign{ & {5 over {3{x^3} – 12x}} = {5 over {3xleft( {x – 2} right)left( {x + 2} right)}} = {{5.2left( {x + 3} right)} over {3xleft( {x – 2} right)left( {x + 2} right).2left( {x + 3} right)}} cr & = {{10left( {x + 3} right)} over {6xleft( {x – 2} right)left( {x + 2} right)left( {x + 3} right)}} cr & {3 over {left( {2x + 4} right)left( {x + 3} right)}} = {3 over {2left( {x + 2} right)left( {x + 3} right)}} = {{3.3xleft( {x – 2} right)} over {2left( {x + 2} right)left( {x + 3} right).3xleft( {x – 2} right)}} cr & = {{9xleft( {x – 2} right)} over {6xleft( {x + 2} right)left( {x – 2} right)left( {x + 3} right)}} cr} )