РЕБЯТА,КТО-НИБУДЬ,ПРОШУ,ПОМОГИТЕ. Created by sinusoida2000. algebra-ru

РЕБЯТА,КТО-НИБУДЬ,ПРОШУ,ПОМОГИТЕ...

sinusoida2000 @sinusoida2000

August 2022 1 24 Report

РЕБЯТА,КТО-НИБУДЬ,ПРОШУ,ПОМОГИТЕ

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NNNLLL54

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$1)\; \; t\cdot \frac{t+ \frac{2}{\sqrt{t+4}} }{2-\sqrt{t+4}} +\sqrt{t+4}+ \frac{4}{\sqrt{t+4}} =t\cdot \frac{t\cdot \sqrt{t+4}+2}{\sqrt{t+4}\cdot (2-\sqrt{t+4})} +\frac{(t+4)+4}{\sqrt{t+4}} =\\\\=\frac{t\cdot (t\cdot \sqrt{t+4}+2)+(t+8)(2-\sqrt{t+4})}{\sqrt{t+4}\cdot (2-\sqrt{t+4})} = \frac{t^2\cdot \sqrt{t+4}+2t+2t+16-t\cdot \sqrt{t+4}-8\cdot \sqrt{t+4}}{\sqrt{t+4}\cdot (2-\sqrt{t+4})} =\\\\= \frac{t^2\cdot \sqrt{t+4}+4t-t\cdot \sqrt{t+4}-8\cdot \sqrt{t+4}+16}{\sqrt{t+4}\cdot (2-\sqrt{t+4})}\; ;$

$2)\; \; t\cdot \frac{1+ \frac{2}{\sqrt{t+4}} }{2-\sqrt{t+4}} +\sqrt{t+4}+ \frac{4}{\sqrt{t+4}} =t\cdot \frac{\sqrt{t+4}+2}{\sqrt{t+4}\cdot (2-\sqrt{t+4})} +\frac{t+8}{\sqrt{t+4}} =\\\\= \frac{t\sqrt{t+4}+2t+(2-\sqrt{t+4})(t+8)}{\sqrt{t+4}\cdot (2-\sqrt{t+4})} = \frac{t\sqrt{t+4}+4t+16-t\sqrt{t+4}-8\sqrt{t+4}}{\sqrt{t+4}(2-\sqrt{t+4})} =\\\\= \frac{4t-8\sqrt{t+4}+16}{\sqrt{t+4}(2-\sqrt{t+4})} =\frac{4(t-2\sqrt{t+4}+4)}{\sqrt{t+4}(2-\sqrt{t+4})}$

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