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Question-208502




Question Number 208502 by Kalebwizeman last updated on 17/Jun/24
Answered by A5T last updated on 17/Jun/24
=(√(16))+(√(15))−(√(15))−(√(14))+...−(√9)−(√8)=4−2(√2)
$$=\sqrt{\mathrm{16}}+\sqrt{\mathrm{15}}−\sqrt{\mathrm{15}}−\sqrt{\mathrm{14}}+…−\sqrt{\mathrm{9}}−\sqrt{\mathrm{8}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by Kalebwizeman last updated on 17/Jun/24
kindly show more working to make it clearer. Thank you
$${kindly}\:{show}\:{more}\:{working}\:{to}\:{make}\:{it}\:{clearer}.\:{Thank}\:{you} \\ $$
Commented by A5T last updated on 17/Jun/24
Rationalize (1/( (√(a+1))−(√a)))=((((√(a+1))+(√a)))/(((√(a+1))−(√a))((√(a+1))+(√a))))  =(((√(a+1))+(√a))/(((√(a+1)))^2 −((√a))^2 ))=(((√(a+1))+(√a))/(a+1−a))=(√(a+1))+(√a)
$${Rationalize}\:\frac{\mathrm{1}}{\:\sqrt{{a}+\mathrm{1}}−\sqrt{{a}}}=\frac{\left(\sqrt{{a}+\mathrm{1}}+\sqrt{{a}}\right)}{\left(\sqrt{{a}+\mathrm{1}}−\sqrt{{a}}\right)\left(\sqrt{{a}+\mathrm{1}}+\sqrt{{a}}\right)} \\ $$$$=\frac{\sqrt{{a}+\mathrm{1}}+\sqrt{{a}}}{\left(\sqrt{{a}+\mathrm{1}}\right)^{\mathrm{2}} −\left(\sqrt{{a}}\right)^{\mathrm{2}} }=\frac{\sqrt{{a}+\mathrm{1}}+\sqrt{{a}}}{{a}+\mathrm{1}−{a}}=\sqrt{{a}+\mathrm{1}}+\sqrt{{a}} \\ $$

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