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 {xleft( {x + 3} ight)}})

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 {xleft( {x + 3} ight)}} = {{x + 1} over {2left( {x + 3} ight)}} + {{2x + 3} over {xleft( {x + 3} ight)}})

( = {{xleft( {x + 1} ight)} over {2xleft( {x + 3} ight)}} + {{2left( {2x + 3} ight)} over {2xleft( {x + 3} ight)}} = {{{x^2} + x + 4x + 6} over {2xleft( {x + 3} ight)}})

( = {{{x^2} + 5x + 6} over {2xleft( {x + 3} ight)}} = {{{x^2} + 2x + 3x + 6} over {2xleft( {x + 3} ight)}} = {{xleft( {x + 2} ight) + 3left( {x + 2} ight)} over {2xleft( {x + 3} ight)}})

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 {xleft( {x – 5} ight)}} + {{x – 25} over {5left( {x – 5} ight)}} = {{5left( {3x + 5} ight)} over {5xleft( {x – 5} ight)}} + {{xleft( {x – 25} ight)} over {5xleft( {x – 5} ight)}})

( = {{15x + 25 + {x^2} – 25x} over {5xleft( {x – 5} ight)}} = {{{x^2} – 10x + 25} over {5xleft( {x – 5} ight)}})

( = {{{{left( {x – 5} ight)}^2}} over {5xleft( {x – 5} ight)}} = {{x – 5} over {5x}})

d)({x^2} + {{{x^4} + 1} over {1 – {x^2}}} + 1 = 1 + {{ m{x}}^2} + {{{x^4} + 1} over {1 – {x^2}}})

( = {{left( {1 + {x^2}} ight)left( {1 – {x^2}} ight)} 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} ight)left( {{x^2} + x + 1} ight)}} + {{2x – 1} over {{x^2} + x + 1}} + {{ – 6} over {x – 1}})

( = {{4{x^2} – 3x + 17} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}} + {{left( {2x – 1} ight)left( {x – 1} ight)} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}} + {{ – 6left( {{x^2} + x + 1} ight)} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}})

( = {{4{x^2} – 3x + 17 + left( {2x – 1} ight)left( {x – 1} ight) – 6left( {{x^2} + x + 1} ight)} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}})

( = {{4{x^2} – 3x + 17 + 2{x^2} – 3x + 1 – 6{x^2} – 6x – 6} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}})

( = {{ – 12x + 12} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}} = {{ – 12left( {x – 1} ight)} over {left( {x – 1} ight)left( {{x^2} + x + 1} ight)}} = {{ – 12} over {{x^2} + x + 1}})