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. \({{4x - 4} \over {2x\left( {x + 3} \right)}},{{x - 3} \over {3x\left( {x + 1} \right)}}\)

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 = \(3x\left( {x + 3} \right)\left( {x + 1} \right)\)  Vì \({{4x - 4} \over {2x\left( {x + 3} \right)}} = {{2\left( {x - 1} \right)} \over {x\left( {x + 3} \right)}}\)

\({{4x - 4} \over {2x\left( {x + 3} \right)}} = {{2\left( {x - 1} \right)} \over {x\left( {x + 3} \right)}} = {{2\left( {x - 1} \right).3\left( {x + 1} \right)} \over {x\left( {x + 3} \right).3\left( {x + 1} \right)}} = {{6\left( {{x^2} - 1} \right)} \over {3x\left( {x + 3} \right)\left( {x + 1} \right)}}\)

\({{x - 3} \over {3x\left( {x + 1} \right)}} = {{\left( {x - 3} \right)\left( {x + 3} \right)} \over {3x\left( {x + 1} \right)\left( {x + 3} \right)}} = {{{x^2} - 9} \over {3x\left( {x + 1} \right)\left( {x + 3} \right)}}\)

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. \(3{x^3} - 12x = 3x\left( {{x^2} - 4} \right) = 3x\left( {x - 2} \right)\left( {x + 2} \right)\)

\(\left( {2x + 4} \right)\left( {x + 3} \right) = 2\left( {x + 2} \right)\left( {x + 3} \right)\)

MTC = \(6x\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 3} \right)\)

\(\eqalign{  & {5 \over {3{x^3} - 12x}} = {5 \over {3x\left( {x - 2} \right)\left( {x + 2} \right)}} = {{5.2\left( {x + 3} \right)} \over {3x\left( {x - 2} \right)\left( {x + 2} \right).2\left( {x + 3} \right)}}  \cr  &  = {{10\left( {x + 3} \right)} \over {6x\left( {x - 2} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}  \cr  & {3 \over {\left( {2x + 4} \right)\left( {x + 3} \right)}} = {3 \over {2\left( {x + 2} \right)\left( {x + 3} \right)}} = {{3.3x\left( {x - 2} \right)} \over {2\left( {x + 2} \right)\left( {x + 3} \right).3x\left( {x - 2} \right)}}  \cr  &  = {{9x\left( {x - 2} \right)} \over {6x\left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 3} \right)}} \cr} \)