Toán Lớp 8: 4/x^2-9+1-x/x^2+3x. Giúp mình với ạ mình đang cần gấp
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Toán Lớp 8: 4/x^2-9+1-x/x^2+3x. Giúp mình với ạ mình đang cần gấp
Comments ( 1 )
DK:x \ne \left\{ { – 3;0;3} \right\}\\
\dfrac{4}{{{x^2} – 9}} + \dfrac{{1 – x}}{{{x^2} + 3x}}\\
= \dfrac{{4x + \left( {1 – x} \right)\left( {x – 3} \right)}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4x – {x^2} + 3x + x – 3}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{ – {x^2} + 8x – 3}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}
\end{array}\)