Bài 21 trang 29 SBT toán 8 tập 1

Đề bài

Làm tính cộng các phân thức 

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

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

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

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

Lời giải

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

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

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

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

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

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

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

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