Bài 25 trang 47 sgk Toán 8 tập 1, Làm tính cộng các phân thức sau:...

Làm tính cộng các phân thức sau:

a)${5 \over {2{x^2}y}} + {3 \over {5x{y^2}}} + {x \over {{y^3}}}$

b)${{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}}$

c)${{3x + 5} \over {{x^2} - 5x}} + {{25 - x} \over {25 - 5x}}$

d)${x^2} + {{{x^4} + 1} \over {1 - {x^2}}} + 1)$

e)${{4{x^2} - 3x + 17} \over {{x^3} - 1}} + {{2x - 1} \over {{x^2} + x + 1}} + {6 \over {1 - x}}$

Hướng dẫn làm bài:

a)${5 \over {2{x^2}y}} + {3 \over {5x{y^2}}} + {x \over {{y^3}}} = {{5.5{y^2}} \over {2{x^2}y.5{y^2}}} + {{3.2xy} \over {5x{y^2}.2xy}} + {{x.10{x^2}} \over {{y^3}.10{x^2}}}$

$= {{25{y^2}} \over {10{x^2}{y^3}}} + {{6xy} \over {10{x^2}{y^3}}} + {{10{x^3}} \over {10{x^2}{y^3}}} = {{25{y^2} + 6xy + 10{x^3}} \over {10{x^2}{y^3}}}$

b)${{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}} = {{x + 1} \over {2\left( {x + 3} \right)}} + {{2x + 3} \over {x\left( {x + 3} \right)}}$

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

$= {{{x^2} + 5x + 6} \over {2x\left( {x + 3} \right)}} = {{{x^2} + 2x + 3x + 6} \over {2x\left( {x + 3} \right)}} = {{x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \over {2x\left( {x + 3} \right)}}$

c)${{3x + 5} \over {{x^2} - 5x}} + {{25 - x} \over {25 - 5x}} = {{3x + 5} \over {{x^2} - 5x}} + {{x - 25} \over {5x - 25}}$

$= {{3x + 5} \over {x\left( {x - 5} \right)}} + {{x - 25} \over {5\left( {x - 5} \right)}} = {{5\left( {3x + 5} \right)} \over {5x\left( {x - 5} \right)}} + {{x\left( {x - 25} \right)} \over {5x\left( {x - 5} \right)}}$

$= {{15x + 25 + {x^2} - 25x} \over {5x\left( {x - 5} \right)}} = {{{x^2} - 10x + 25} \over {5x\left( {x - 5} \right)}}$

$= {{{{\left( {x - 5} \right)}^2}} \over {5x\left( {x - 5} \right)}} = {{x - 5} \over {5x}}$

d)${x^2} + {{{x^4} + 1} \over {1 - {x^2}}} + 1 = 1 + {{\rm{x}}^2} + {{{x^4} + 1} \over {1 - {x^2}}}$

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

e)${{4{x^2} - 3x + 17} \over {{x^3} - 1}} + {{2x - 1} \over {{x^2} + x + 1}} + {6 \over {1 - x}}$

${{4{x^2} - 3x + 17} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{2x - 1} \over {{x^2} + x + 1}} + {{ - 6} \over {x - 1}}$

$= {{4{x^2} - 3x + 17} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{\left( {2x - 1} \right)\left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + {{ - 6\left( {{x^2} + x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$

$= {{4{x^2} - 3x + 17 + \left( {2x - 1} \right)\left( {x - 1} \right) - 6\left( {{x^2} + x + 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$

$= {{4{x^2} - 3x + 17 + 2{x^2} - 3x + 1 - 6{x^2} - 6x - 6} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$

$= {{ - 12x + 12} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{ - 12\left( {x - 1} \right)} \over {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = {{ - 12} \over {{x^2} + x + 1}}$