Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 58390 by MJS last updated on 22/Apr/19

$$\mathrm{write}\mathrm{without}\mathrm{roots}\mathrm{in}\mathrm{denominator}\mathrm{if}\mathrm{possible}$$$$\left(1\right)\frac{1}{\sqrt{a}}$$$$\left(2\right)\frac{1}{\sqrt{a}+\sqrt{b}}$$$$\left(3\right)\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$$$$\left(4\right)\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}$$$$\left(5\right)\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e}}$$

Commented by tanmay last updated on 22/Apr/19

$$sirplscheckmyanswer...$$

Commented by MJS last updated on 22/Apr/19

$$\mathrm{your}\mathrm{answers}\mathrm{are}\mathrm{right},\mathrm{thank}\mathrm{you}$$$$\mathrm{I}\mathrm{think}\mathrm{the}{5}^{\mathrm{th}}\mathrm{one}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{possible}\mathrm{but}{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{not}$$$$\mathrm{sure}.$$

Answered by tanmay last updated on 22/Apr/19

$$1)\frac{\sqrt{a}}{a}$$$$2)\frac{\sqrt{a}-\sqrt{b}}{a-b}$$$$3)\frac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{{(\sqrt{a}+\sqrt{b})}^2-c}$$$$=\frac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{a+b-c+2\sqrt{a}\sqrt{b}}$$$$=\frac{(\sqrt{a}+\sqrt{b}-\sqrt{c})(a+b-c-2\sqrt{a}\sqrt{b})}{{(a+b-c)}^2-4ab}$$

Answered by tanmay last updated on 22/Apr/19

$$4)\frac{(\sqrt{a}+\sqrt{b})-(\sqrt{c}+\sqrt{d})}{a+b-c-d+2\sqrt{a}\sqrt{b}-2\sqrt{c}\sqrt{d}}$$$$\frac{\{(\sqrt{a}+\sqrt{b})-(\sqrt{c}+\sqrt{d}\}\{a+b-c-d-2\sqrt{a}\sqrt{b}+2\sqrt{cd}\}}{\{{(a+b-c-d)}^2-4ab-4cd\}+8\sqrt{abcd}}$$$$=\frac{\{(\sqrt{a}+\sqrt{b})-(\sqrt{c}+\sqrt{d})\}\{a+b-c-d-2\sqrt{ab}+2\sqrt{cd}\}[\{{(a+b-c-d)}^2-4ab-4cd\}-8\sqrt{abcd}\}}{{\{{(a+b-c-d)}^2-4ab-4cd\}}^2-64abcd}$$$$$$

Commented by malwaan last updated on 23/Apr/19

$$trytosolve5$$$$please$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com