Toán Lớp 8: a, $\frac{1}{x+2}$ , $\frac{8}{2x-x^2}$
b, $\frac{x^4}{x^2+1}$ , x^2+1
c, $\frac{3}{x^3-3x^2y+3xy^2-y^3}$ , $\frac{x}{y^2-xy}$
d, $\frac{25}{14x^2y}$ , $\frac{14}{21xy^5}$
e, $\frac{4x-4}{2x(x+3)}$ , $\frac{x-3}{3x(x+1)}$
h, $\frac{5}{3x^3-12x}$ , $\frac{3}{(2x+4)(x+3)}$
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a)2x – {x^2} = x.\left( {2 – x} \right)\\
\Leftrightarrow MSC = x.\left( {2 – x} \right).\left( {x + 2} \right)\\
+ \dfrac{1}{{x + 2}} = \dfrac{{x\left( {2 – x} \right)}}{{x.\left( {2 – x} \right).\left( {x + 2} \right)}} = \dfrac{{2x – {x^2}}}{{x.\left( {2 – x} \right).\left( {x + 2} \right)}}\\
+ \dfrac{8}{{2x – {x^2}}} = \dfrac{{8.\left( {x + 2} \right)}}{{x.\left( {2 – x} \right).\left( {x + 2} \right)}} = \dfrac{{8x + 16}}{{x.\left( {2 – x} \right).\left( {x + 2} \right)}}\\
b)MSC = {x^2} + 1\\
+ \dfrac{{{x^4}}}{{{x^2} + 1}}\\
+ {x^2} + 1 = \dfrac{{\left( {{x^2} + 1} \right).\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right)}} = \dfrac{{{x^4} + 2{x^2} + 1}}{{\left( {{x^2} + 1} \right)}}\\
c)\\
{x^3} – 3{x^2}y + 3x{y^2} – {y^3} = {\left( {x – y} \right)^3}\\
{y^2} – xy = y.\left( {y – x} \right)\\
\Leftrightarrow MSC = y.{\left( {x – y} \right)^3}\\
+ \dfrac{3}{{{x^3} – 3{x^2}y + 3x{y^2} – {y^3}}} = \dfrac{{3.y}}{{y.{{\left( {x – y} \right)}^3}}}\\
+ \dfrac{x}{{{y^2} – xy}} = \dfrac{{ – x.{{\left( {x – y} \right)}^2}}}{{y.{{\left( {x – y} \right)}^3}}} = \dfrac{{ – {x^3} + 2{x^2}y – x{y^2}}}{{y.{{\left( {x – y} \right)}^3}}}\\
d)14{x^2}y;21x{y^5}\\
\Leftrightarrow MSC = 42{x^2}{y^5}\\
+ \dfrac{{25}}{{14{x^2}y}} = \dfrac{{25.3{y^4}}}{{42{x^2}{y^5}}} = \dfrac{{75{y^4}}}{{42{x^2}{y^5}}}\\
+ \dfrac{{14}}{{21x{y^5}}} = \dfrac{{14.2x}}{{42{x^2}{y^5}}} = \dfrac{{28x}}{{42{x^2}{y^5}}}\\
e)MSC = 6x\left( {x + 3} \right)\left( {x + 1} \right)\\
+ \dfrac{{4x – 4}}{{2x\left( {x + 3} \right)}} = \dfrac{{\left( {4x – 4} \right).3\left( {x + 1} \right)}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}}\\
= \dfrac{{12\left( {{x^2} – 1} \right)}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}} = \dfrac{{12{x^2} – 12}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}}\\
+ \dfrac{{x – 3}}{{3x\left( {x + 1} \right)}} = \dfrac{{\left( {x – 3} \right).2\left( {x + 3} \right)}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2\left( {{x^2} – 9} \right)}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}} = \dfrac{{2{x^2} – 18}}{{6x\left( {x + 3} \right)\left( {x + 1} \right)}}\\
h)3{x^3} – 12x = 3x\left( {{x^2} – 4} \right) = 3x\left( {x – 2} \right)\left( {x + 2} \right)\\
\left( {2x + 4} \right)\left( {x + 3} \right) = 2\left( {x + 2} \right)\left( {x + 3} \right)\\
\Leftrightarrow MSC = 6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)\\
+ )\dfrac{5}{{3{x^3} – 12x}} = \dfrac{{5.2\left( {x + 3} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{10x + 30}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
+ )\dfrac{3}{{\left( {2x + 4} \right)\left( {x + 3} \right)}} = \dfrac{{3.3x\left( {x – 2} \right)}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{9{x^2} – 18x}}{{6x\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 3} \right)}}
\end{array}$