Câu 21 trang 29 Sách bài tập (SBT) Toán 8 tập 1

Làm tính cộng các phân thức

a. \({{11x + 13} \over {3x - 3}} + {{15x + 17} \over {4 - 4x}}\)

b. \({{2x + 1} \over {2{x^2} - x}} + {{32{x^2}} \over {1 - 4{x^2}}} + {{1 - 2x} \over {2{x^2} + x}}\)

c. \({1 \over {{x^2} + x + 1}} + {1 \over {{x^2} - x}} + {{2x} \over {1 - {x^3}}}\)

d. \({{{x^4}} \over {1 - x}} + {x^3} + {x^2} + x + 1\)

Giải:

a. \({{11x + 13} \over {3x - 3}} + {{15x + 17} \over {4 - 4x}}\)\( = {{11x + 13} \over {3\left( {x - 1} \right)}} + {{ - 15x - 17} \over {4\left( {x - 1} \right)}}\)

\(\eqalign{  &  = {{4\left( {11x + 13} \right)} \over {12\left( {x - 1} \right)}} + {{3\left( { - 15x - 17} \right)} \over {12\left( {x - 1} \right)}} = {{44x + 52 - 45x - 51} \over {12\left( {x - 1} \right)}} = {{1 - x} \over {12\left( {x - 1} \right)}}  \cr  &  = {{ - \left( {x - 1} \right)} \over {12\left( {x - 1} \right)}} =  - {1 \over {12}} \cr} \)

b. \({{2x + 1} \over {2{x^2} - x}} + {{32{x^2}} \over {1 - 4{x^2}}} + {{1 - 2x} \over {2{x^2} + x}}\)\( = {{2x + 1} \over {x\left( {2x - 1} \right)}} + {{ - 32{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} + {{1 - 2x} \over {x\left( {2x + 1} \right)}}\)

\(\eqalign{  &  = {{\left( {2x + 1} \right)\left( {2x + 1} \right)} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}} + {{ - 32{x^2}.x} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}} + {{\left( {1 - 2x} \right)\left( {2x - 1} \right)} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}}  \cr  &  = {{4{x^2} + 4x + 1 - 32{x^3} + 2x - 1 - 4{x^2} + 2x} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}} = {{ - 32{x^3} + 8x} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}}  \cr  &  = {{ - 8x\left( {4{x^2} - 1} \right)} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}} = {{ - 8x\left( {2x + 1} \right)\left( {2x - 1} \right)} \over {x\left( {2x + 1} \right)\left( {2x - 1} \right)}} =  - 8 \cr} \)

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

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

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

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

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