Toán Lớp 8: Tìm x :
x – $\frac{x^{2}+1}{x-1}$ + 1 = $\frac{x-10}{2-x}$ – $\frac{x-5}{2-x}$
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Toán Lớp 8: Tìm x :
x – $\frac{x^{2}+1}{x-1}$ + 1 = $\frac{x-10}{2-x}$ – $\frac{x-5}{2-x}$
Comments ( 1 )
DK:x \ne \left\{ {1;2} \right\}\\
x – \dfrac{{{x^2} + 1}}{{x – 1}} + 1 = \dfrac{{x – 10}}{{2 – x}} – \dfrac{{x – 5}}{{2 – x}}\\
\to \dfrac{{x\left( {x – 1} \right) – {x^2} – 1 + x – 1}}{{x – 1}} = \dfrac{{x – 10 – x + 5}}{{2 – x}}\\
\to \dfrac{{{x^2} – x – {x^2} + x – 2}}{{x – 1}} = \dfrac{{ – 5}}{{2 – x}}\\
\to \dfrac{{ – 2}}{{x – 1}} = \dfrac{5}{{x – 2}}\\
\to – 2x + 4 = 5x – 5\\
\to 7x = 9\\
\to x = \dfrac{9}{7}
\end{array}\)