Thực hiện phép tính
a) $\frac{x^2-x-2}{x^2+x-6}$ . $\frac{x^2-x-12}{x^2+6x+5}$
b) $\frac{x^2-49}{x^2+2x+1}$ . $\frac{x^2+4x-21}{x^2-2x-3}$
c) $\frac{x^3+y^3}{x^2y^2+2xy^3+y^4}$ : $\frac{x^2-xy}{x^2-y^2}$
Thực hiện phép tính
a) $\frac{x^2-x-2}{x^2+x-6}$ . $\frac{x^2-x-12}{x^2+6x+5}$
b) $\frac{x^2-49}{x^2+2x+1}$ . $\frac{x^2+4x-21}{x^2-2x-3}$
c) $\frac{x^3+y^3}{x^2y^2+2xy^3+y^4}$ : $\frac{x^2-xy}{x^2-y^2}$
Đáp án:
c) \(\dfrac{{{x^2} – xy + {y^2}}}{{x{y^2}}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^2} + x – 2x – 2}}{{{x^2} + 3x – 2x – 6}}.\dfrac{{{x^2} – 4x + 3x – 12}}{{{x^2} + x + 5x + 5}}\\
= \dfrac{{x\left( {x + 1} \right) – 2\left( {x + 1} \right)}}{{x\left( {x + 3} \right) – 2\left( {x + 3} \right)}}.\dfrac{{x\left( {x – 4} \right) + 3\left( {x – 4} \right)}}{{x\left( {x + 1} \right) + 5\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x – 2} \right)}}{{\left( {x + 3} \right)\left( {x – 2} \right)}}.\dfrac{{\left( {x – 4} \right)\left( {x + 3} \right)}}{{\left( {x + 1} \right)\left( {x + 5} \right)}}\\
= \dfrac{{x – 4}}{{x + 5}}\\
b)\dfrac{{\left( {x – 7} \right)\left( {x + 7} \right)}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{{x^2} + 7x – 3x – 21}}{{{x^2} + x – 3x – 3}}\\
= \dfrac{{\left( {x – 7} \right)\left( {x + 7} \right)}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{x\left( {x + 7} \right) – 3\left( {x + 7} \right)}}{{x\left( {x + 1} \right) – 3\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x – 7} \right)\left( {x + 7} \right)}}{{{{\left( {x + 1} \right)}^2}}}.\dfrac{{\left( {x + 7} \right)\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x – 7} \right){{\left( {x + 7} \right)}^2}}}{{{{\left( {x + 1} \right)}^3}}}\\
c)\dfrac{{\left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right)}}{{{y^2}\left( {{x^2} + 2xy + {y^2}} \right)}}.\dfrac{{\left( {x – y} \right)\left( {x + y} \right)}}{{x\left( {x – y} \right)}}\\
= \dfrac{{\left( {x + y} \right)\left( {{x^2} – xy + {y^2}} \right)}}{{{y^2}{{\left( {x + y} \right)}^2}}}.\dfrac{{\left( {x – y} \right)\left( {x + y} \right)}}{{x\left( {x – y} \right)}}\\
= \dfrac{{{x^2} – xy + {y^2}}}{{x{y^2}}}
\end{array}\)
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