Bài 19 trang 43 SGK Toán 8 tập 1

Quy đồng mẫu thức các phân thức sau:

a) \(\dfrac{1}{{x + 2}},\dfrac{8}{{2x - {x^2}}}\)

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

c) \(\dfrac{{{x^3}}}{{{x^3} - 3{x^2}y + 3x{y^2} - {y^3}}},\dfrac{x}{{{y^2} - xy}}\)

a) MTC = \(x\left( {2 - x} \right)\left( {2 + x} \right)\)

\(\dfrac{1}{{x + 2}} = \dfrac{1}{{2 + x}} = \dfrac{{x\left( {2 - x} \right)}}{{x\left( {2 - x} \right)\left( {2 + x} \right)}}\)\(\, = \dfrac{{2x - {x^2}}}{{x(2 - x)(2 + x)}}\)

\(\dfrac{8}{{2x - {x^2}}} = \dfrac{{8.(2 + x)}}{{x(2 - x)(2 + x)}}\)\(\, = \dfrac{{16 + 8x}}{{x(2 - x)(2 + x)}}\)

b) MTC = \({x^2} - 1\)

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

\(\dfrac{{{x^4}}}{{{x^2} - 1}}\) giữ nguyên.

c) Ta có: \({x^3} - 3{x^2}y + 3x{y^2} - {y^3} = {\left( {x - y} \right)^3}\)

\({y^2} - xy = y\left( {y - x} \right) =  - y\left( {x - y} \right)\)

 MTC = \(y{\left( {x - y} \right)^3}\)

+ Quy đồng mẫu thức :

\(\dfrac{{{x^3}}}{{{x^3} - 3{x^2}y + 3x{y^2} - {y^3}}} = \dfrac{{{x^3}}}{{{{\left( {x - y} \right)}^3}}} \)\(\,= \dfrac{{{x^3}y}}{{y{{\left( {x - y} \right)}^3}}}\)

\(\dfrac{x}{{{y^2} - xy}} = \dfrac{x}{{y\left( {y - x} \right)}} = \dfrac{x}{{ - y\left( {x - y} \right)}}\)\(\, = \dfrac{{ - x}}{{y\left( {x - y} \right)}} = \dfrac{{ - x{{\left( {x - y} \right)}^2}}}{{y{{(x - y)}^3}}}\)