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

write without roots in denominator if possible (1) (1/(√a)) (2) (1/((√a)+(√b))) (3) (1/((√a)+(√b)+(√c))) (4) (1/((√a)+(√b)+(√c)+(√d))) (5) (1/((√a)+(√b)+(√c)+(√d)+(√e)))

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

Commented by tanmay last updated on 22/Apr/19

sir pls check my answer...

$${sir}\:{pls}\:{check}\:{my}\:{answer}... \\ $$

Commented by MJS last updated on 22/Apr/19

your answers are right, thank you I think the 5^(th) one isn′t possible but I′m not sure.

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

Answered by tanmay last updated on 22/Apr/19

1)((√a)/a) 2)(((√a)−(√b))/(a−b)) 3)(((√a) +(√b) −(√c))/(((√a) +(√b) )^2 −c)) =(((√a) +(√b) −(√c))/(a+b−c+2(√a) (√b))) =((((√a) +(√b) −(√c) )(a+b−c−2(√a) (√b) ))/((a+b−c)^2 −4ab))

$$\left.\mathrm{1}\right)\frac{\sqrt{{a}}}{{a}} \\ $$$$\left.\mathrm{2}\right)\frac{\sqrt{{a}}−\sqrt{{b}}}{{a}−{b}} \\ $$$$\left.\mathrm{3}\right)\frac{\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}}{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)^{\mathrm{2}} −{c}} \\ $$$$=\frac{\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}}{{a}+{b}−{c}+\mathrm{2}\sqrt{{a}}\:\sqrt{{b}}}\: \\ $$$$=\frac{\left(\sqrt{{a}}\:+\sqrt{{b}}\:−\sqrt{{c}}\:\right)\left({a}+{b}−{c}−\mathrm{2}\sqrt{{a}}\:\sqrt{{b}}\:\right)}{\left({a}+{b}−{c}\right)^{\mathrm{2}} −\mathrm{4}{ab}} \\ $$

Answered by tanmay last updated on 22/Apr/19

4)((((√a) +(√b) )−((√c) +(√d) ))/(a+b−c−d+2(√a) (√(b )) −2(√c) (√d))) (({((√a) +(√b) )−((√c) +(√d) }{a+b−c−d−2(√a) (√b) +2(√(cd)) })/({(a+b−c−d)^2 −4ab−4cd}+8(√(abcd)) )) =(({((√a) +(√b) )−((√c) +(√d) )}{a+b−c−d−2(√(ab)) +2(√(cd)) }[{(a+b−c−d)^2 −4ab−4cd}−8(√(abcd)) })/({(a+b−c−d)^2 −4ab−4cd}^2 −64abcd))

$$\left.\mathrm{4}\right)\frac{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)−\left(\sqrt{{c}}\:+\sqrt{{d}}\:\right)}{{a}+{b}−{c}−{d}+\mathrm{2}\sqrt{{a}}\:\sqrt{{b}\:}\:−\mathrm{2}\sqrt{{c}}\:\sqrt{{d}}} \\ $$$$\frac{\left\{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)−\left(\sqrt{{c}}\:+\sqrt{{d}}\:\right\}\left\{{a}+{b}−{c}−{d}−\mathrm{2}\sqrt{{a}}\:\sqrt{{b}}\:+\mathrm{2}\sqrt{{cd}}\:\right\}\right.}{\left\{\left({a}+{b}−{c}−{d}\right)^{\mathrm{2}} −\mathrm{4}{ab}−\mathrm{4}{cd}\right\}+\mathrm{8}\sqrt{{abcd}}\:}\: \\ $$$$=\frac{\left\{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)−\left(\sqrt{{c}}\:+\sqrt{{d}}\:\right)\right\}\left\{{a}+{b}−{c}−{d}−\mathrm{2}\sqrt{{ab}}\:+\mathrm{2}\sqrt{{cd}}\:\right\}\left[\left\{\left({a}+{b}−{c}−{d}\right)^{\mathrm{2}} −\mathrm{4}{ab}−\mathrm{4}{cd}\right\}−\mathrm{8}\sqrt{{abcd}}\:\right\}}{\left\{\left({a}+{b}−{c}−{d}\right)^{\mathrm{2}} −\mathrm{4}{ab}−\mathrm{4}{cd}\right\}^{\mathrm{2}} −\mathrm{64}{abcd}} \\ $$$$ \\ $$

Commented by malwaan last updated on 23/Apr/19

try to solve 5 please

$${try}\:{to}\:{solve}\:\mathrm{5} \\ $$$${please} \\ $$

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