Toán Lớp 8: cho P = ( $\frac{x}{x^{2}-36}$ – $\frac{x-6}{x^{2}+6x}$ ) : $\frac{2x-6}{x^{2}+6x}$ + $\frac{x}{6-x}$ – $\frac{9}{x^{2}+5}$
Tìm x khi P = 4
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Toán Lớp 8: cho P = ( $\frac{x}{x^{2}-36}$ – $\frac{x-6}{x^{2}+6x}$ ) : $\frac{2x-6}{x^{2}+6x}$ + $\frac{x}{6-x}$ – $\frac{9}{x^{2}+5}$
Tìm x khi P = 4
Comments ( 1 )
DK:x \ne \left\{ { – 6;0;6} \right\}\\
P = \left( {\dfrac{x}{{{x^2} – 36}} – \dfrac{{x – 6}}{{{x^2} + 6x}}} \right):\dfrac{{2x – 6}}{{{x^2} + 6x}} + \dfrac{x}{{6 – x}} – \dfrac{9}{{{x^2} + 5}}\\
= \dfrac{{{x^2} – {{\left( {x – 6} \right)}^2}}}{{x\left( {x + 6} \right)\left( {x – 6} \right)}}.\dfrac{{x\left( {x + 6} \right)}}{{2\left( {x – 3} \right)}} + \dfrac{x}{{6 – x}} – \dfrac{9}{{{x^2} + 5}}\\
= \dfrac{{{x^2} – {x^2} + 12x – 36}}{{x\left( {x + 6} \right)\left( {x – 6} \right)}}.\dfrac{{x\left( {x + 6} \right)}}{{2\left( {x – 3} \right)}} – \dfrac{x}{{x – 6}} – \dfrac{9}{{{x^2} + 5}}\\
= \dfrac{{12\left( {x – 3} \right)}}{{x\left( {x + 6} \right)\left( {x – 6} \right)}}.\dfrac{{x\left( {x + 6} \right)}}{{2\left( {x – 3} \right)}} – \dfrac{x}{{x – 6}} – \dfrac{9}{{{x^2} + 5}}\\
= \dfrac{6}{{x – 6}} – \dfrac{x}{{x – 6}} – \dfrac{9}{{{x^2} + 5}}\\
= \dfrac{{6 – x}}{{x – 6}} – \dfrac{9}{{{x^2} + 5}}\\
= – 1 – \dfrac{9}{{{x^2} + 5}}\\
P = 4\\
\to – 1 – \dfrac{9}{{{x^2} + 5}} = 4\\
\to \dfrac{9}{{{x^2} + 5}} = – 5\\
\to {x^2} + 5 = – \dfrac{9}{5}\\
\to {x^2} = – \dfrac{{34}}{5}\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)