a) $\frac{1}{9x-18}$ + $\frac{22-7x}{72-18x^2}$ + $\frac{5}{12x+24}$ b)$\frac{x+1}{x-1}$ – $\frac{x-1}{x+1}$ – $\frac{4}{1-x^2}$

Question

a) $\frac{1}{9x-18}$ + $\frac{22-7x}{72-18x^2}$ + $\frac{5}{12x+24}$
b)$\frac{x+1}{x-1}$ – $\frac{x-1}{x+1}$ – $\frac{4}{1-x^2}$

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2 ngày 2021-12-07T19:22:55+00:00 1 Answers 2 views 0

$\begin{array}{l} a)Dkxd:x \ne 2;x \ne – 2\\ \dfrac{1}{{9x – 18}} + \dfrac{{22 – 7x}}{{72 – 18{x^2}}} + \dfrac{5}{{12x + 24}}\\ = \dfrac{1}{{9\left( {x – 2} \right)}} + \dfrac{{22 – 7x}}{{18.\left( {4 – {x^2}} \right)}} + \dfrac{5}{{12.\left( {x + 2} \right)}}\\ = \dfrac{{4.\left( {x + 2} \right) + \left( {7x – 22} \right).2 + 3.\left( {x – 2} \right).5}}{{36\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{4x + 8 + 14x – 44 + 15x – 30}}{{36\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{33x – 66}}{{36\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{33.\left( {x – 2} \right)}}{{36\left( {x – 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{11}}{{12\left( {x + 2} \right)}}\\ b)Dkxd:x \ne 1;x \ne – 1\\ \dfrac{{x + 1}}{{x – 1}} – \dfrac{{x – 1}}{{x + 1}} – \dfrac{4}{{1 – {x^2}}}\\ = \dfrac{{{{\left( {x + 1} \right)}^2} – {{\left( {x – 1} \right)}^2} + 4}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \dfrac{{{x^2} + 2x + 1 – {x^2} + 2x – 1 + 4}}{{\left( {x + 1} \right)\left( {x – 1} \right)}}\\ = \dfrac{{4x + 4}}{{\left( {x + 1} \right)\left( {x – 1} \right)}}\\ = \dfrac{{4\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x – 1} \right)}}\\ = \dfrac{4}{{x – 1}} \end{array}$