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Question Number 200618 by sonukgindia last updated on 21/Nov/23

Commented by Frix last updated on 21/Nov/23

(√2)−1

$$\sqrt{\mathrm{2}}−\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 22/Nov/23

(((√1) +(√2) +(√3) +(√4) +(√5) )/(2(√2) +(√3) +(√4)+(√5) +(√6) +(√8) +(√9) +(√(10)) )) =(((√1) +(√2) +(√3) +(√4) +(√5) )/( (√1)+(√2) +(√3) +(√4)+(√5) +(√2) −(√1) +(√2) ((√3) +(√4) +(√5) )+(√9) )) let (√1) +(√2) +(√3) +(√4) +(√5) =a⇒(√3) +(√4) +(√5) =a−(√1) −(√2) =((a )/( a +(√2) −(√1) +(√2) (a−(√1) −(√2) )+(√9) )) =(a/(a+(√2) −(√1) +(a−1)(√2) −2+3)) =(a/(a+(√2) −1 +(a−1)(√2) +1)) =(a/(a+(√2) +(a−1)(√2) )) =(a/(a +a(√2) ))=(1/(1+(√2)))∙((1−(√2))/(1−(√2)))=((1−(√2))/(−1))=−1+(√2)

$$\frac{\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:+\sqrt{\mathrm{5}}\:}{\mathrm{2}\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}+\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{8}}\:+\sqrt{\mathrm{9}}\:+\sqrt{\mathrm{10}}\:} \\ $$$$=\frac{\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:+\sqrt{\mathrm{5}}\:}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}+\sqrt{\mathrm{5}}\:+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:\:+\sqrt{\mathrm{5}}\:\right)+\sqrt{\mathrm{9}}\:} \\ $$$$\: \\ $$$${let}\:\:\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:+\sqrt{\mathrm{5}}\:={a}\Rightarrow\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{4}}\:\:+\sqrt{\mathrm{5}}\:={a}−\sqrt{\mathrm{1}}\:−\sqrt{\mathrm{2}}\: \\ $$$$\: \\ $$$$=\frac{{a}\:}{\:{a}\:+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{1}}\:+\sqrt{\mathrm{2}}\:\left({a}−\sqrt{\mathrm{1}}\:−\sqrt{\mathrm{2}}\:\:\right)+\sqrt{\mathrm{9}}\:} \\ $$$$=\frac{{a}}{{a}+\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{1}}\:+\left({a}−\mathrm{1}\right)\sqrt{\mathrm{2}}\:\:−\mathrm{2}+\mathrm{3}} \\ $$$$=\frac{{a}}{{a}+\sqrt{\mathrm{2}}\:−\mathrm{1}\:+\left({a}−\mathrm{1}\right)\sqrt{\mathrm{2}}\:\:+\mathrm{1}} \\ $$$$=\frac{{a}}{{a}+\sqrt{\mathrm{2}}\:\:+\left({a}−\mathrm{1}\right)\sqrt{\mathrm{2}}\:\:} \\ $$$$=\frac{{a}}{{a}\:+{a}\sqrt{\mathrm{2}}\:\:}=\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}\centerdot\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{\mathrm{1}−\sqrt{\mathrm{2}}}=\frac{\mathrm{1}−\sqrt{\mathrm{2}}}{−\mathrm{1}}=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$

Commented by Frix last updated on 22/Nov/23

Yes!

$$\mathrm{Yes}! \\ $$

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