Bài 25 trang 47 SGK Toán 8 tập 1

Làm tính cộng các phân thức sau:

a) \(\dfrac{5}{{2{x^2}y}} + \dfrac{3}{{5x{y^2}}} + \dfrac{x}{{{y^3}}}\)

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

c) \(\dfrac{{3x + 5}}{{{x^2} - 5x}} + \dfrac{{25 - x}}{{25 - 5x}}\)

d) \({x^2} + \dfrac{{{x^4} + 1}}{{1 - {x^2}}} + 1\)

e) \(\dfrac{{4{x^2} - 3x + 17}}{{{x^3} - 1}} + \dfrac{{2x - 1}}{{{x^2} + x + 1}} + \dfrac{6}{{1 - x}}\)

\(\eqalign{
& a)\,\,MTC = 10{x^2}{y^3} \cr
& {5 \over {2{x^2}y}} + {3 \over {5x{y^2}}} + {x \over {{y^3}}} \cr
& = {{5.5{y^2}} \over {2{x^2}y.5{y^2}}} + {{3.2xy} \over {5x{y^2}.2xy}} + {{x.10{x^2}} \over {{y^3}.10{x^2}}} \cr
& = {{25{y^2}} \over {10{x^2}{y^3}}} + {{6xy} \over {10{x^2}{y^3}}} + {{10{x^3}} \over {10{x^2}{y^3}}} \cr
& = {{25{y^2} + 6xy + 10{x^3}} \over {10{x^2}{y^3}}} \cr} \)

\(\eqalign{
& b)\,\,MTC = 2x\left( {x + 3} \right) \cr
& {{x + 1} \over {2x + 6}} + {{2x + 3} \over {x\left( {x + 3} \right)}} \cr
& = {{x + 1} \over {2\left( {x + 3} \right)}} + {{2x + 3} \over {x\left( {x + 3} \right)}} \cr
& = {{x\left( {x + 1} \right)} \over {2x\left( {x + 3} \right)}} + {{2\left( {2x + 3} \right)} \over {2x\left( {x + 3} \right)}} \cr
& = {{{x^2} + x} \over {2x\left( {x + 3} \right)}} + {{4x + 6} \over {2x\left( {x + 3} \right)}} \cr
& = {{{x^2} + x + 4x + 6} \over {2x\left( {x + 3} \right)}} \cr
& = {{{x^2} + 5x + 6} \over {2x\left( {x + 3} \right)}} \cr
& = {{{x^2} + 2x + 3x + 6} \over {2x\left( {x + 3} \right)}} \cr
& = {{x\left( {x + 2} \right) + 3\left( {x + 2} \right)} \over {2x\left( {x + 3} \right)}} \cr
& = {{\left( {x + 2} \right)\left( {x + 3} \right)} \over {2x\left( {x + 3} \right)}} = {{x + 2} \over {2x}} \cr} \)

\(\eqalign{
& c)\,\,MTC = 5x\left( {x - 5} \right) \cr
& {{3x + 5} \over {{x^2} - 5x}} + {{25 - x} \over {25 - 5x}} \cr
& = {{3x + 5} \over {{x^2} - 5x}} + {{ - \left( {25 - x} \right)} \over { - \left( {25 - 5x} \right)}}\cr& = {{3x + 5} \over {{x^2} - 5x}} + {{x - 25} \over {5x - 25}} \cr
& = {{3x + 5} \over {x\left( {x - 5} \right)}} + {{x - 25} \over {5\left( {x - 5} \right)}} \cr
& = {{5\left( {3x + 5} \right)} \over {5x\left( {x - 5} \right)}} + {{x\left( {x - 25} \right)} \over {5x\left( {x - 5} \right)}} \cr
& = {{15x + 25} \over {5x\left( {x - 5} \right)}} + {{{x^2} - 25x} \over {5x\left( {x - 5} \right)}} \cr
& = {{15x + 25 + {x^2} - 25x} \over {5x\left( {x - 5} \right)}} \cr
& = {{{x^2} - 10x + 25} \over {5x\left( {x - 5} \right)}} \cr
& = {{{x^2} - 2.x.5 + {5^2}} \over {5x\left( {x - 5} \right)}} \cr
& = {{{{\left( {x - 5} \right)}^2}} \over {5x\left( {x - 5} \right)}} = {{x - 5} \over {5x}} \cr} \)

\(\eqalign{
& d)\,MTC = 1 - {x^2} \cr
& {x^2} + {{{x^4} + 1} \over {1 - {x^2}}} + 1 \cr
& = 1 + {{\rm{x}}^2} + {{{x^4} + 1} \over {1 - {x^2}}} \cr
& = {{\left( {1 + {x^2}} \right)\left( {1 - {x^2}} \right)} \over {1 - {x^2}}} + {{{x^4} + 1} \over {1 - {x^2}}} \cr
& = {{1 - {x^4}} \over {1 - {x^2}}} + {{{x^4} + 1} \over {1 - {x^2}}} \cr
& = {{1 - {x^4} + {x^4} + 1} \over {1 - {x^2}}} = {2 \over {1 - {x^2}}} \cr} \)