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

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. \(2{x^2} + 6x = 2x\left( {x + 3} \right);{x^2} - 9 = \left( {x + 3} \right)\left( {x - 3} \right)\)  MTC = \(2x\left( {x + 3} \right)\left( {x - 3} \right)\)

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

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

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

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

c. \({x^3} - 1 = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\) MTC = \({x^3} - 1\)

\({{4{x^2} - 3x + 5} \over {{x^3} - 1}}\);

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

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

MTC = \(10x\left( {2y + x} \right)\left( {2y - x} \right)\)

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

e.

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

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

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

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