Toán Lớp 8: Thực hiện phép tính : x^2+3x/x^2+6x+9 + 3/x-3 – 6x/x^2-9
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Toán Lớp 8: Thực hiện phép tính : x^2+3x/x^2+6x+9 + 3/x-3 – 6x/x^2-9
ai giải đc mình cho lời giải hay nhát
Comments ( 2 )
DK:x \ne \pm 3\\
\dfrac{{{x^2} + 3x}}{{{x^2} + 6x + 9}} + \dfrac{3}{{x – 3}} – \dfrac{{6x}}{{{x^2} – 9}}\\
= \dfrac{{x\left( {x + 3} \right)}}{{{{\left( {x + 3} \right)}^2}}} + \dfrac{3}{{x – 3}} – \dfrac{{6x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{x}{{x + 3}} + \dfrac{3}{{x – 3}} – \dfrac{{6x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x\left( {x – 3} \right) + 3\left( {x + 3} \right) – 6x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} – 3x + 3x + 9 – 6x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} – 6x + 9}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{{\left( {x – 3} \right)}^2}}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x – 3}}{{x + 3}}
\end{array}\)