Question

Find the value of $\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}+(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})÷\sqrt{\frac{16}{81}}$

• A. $\frac{69}{16}$
• B. $\frac{54}{7}$
• C. $\frac{31}{3}$
• D. $\frac{108}{13}$

Answer 1

The correct option is A $\frac{69}{16}$

Given that

$\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}+(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})÷\sqrt{\frac{16}{81}}$

we can find root step by step

$\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}+(\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}})÷\sqrt{\frac{16}{81}}$

$=\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+15}}}}+(\sqrt{\frac{225}{729}}-\frac{5}{12})÷\frac{4}{9}$

$=\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{169}}}}+(\sqrt{\frac{25}{81}}-\frac{5}{12})÷\frac{4}{9}$

$=\sqrt{10+\sqrt{25+\sqrt{108+13}}}+(\frac{5}{9}-\frac{5}{12})÷\frac{4}{9}$

$=\sqrt{10+\sqrt{25+\sqrt{121}}}+\frac{5}{36}÷\frac{4}{9}$

$=\sqrt{10+\sqrt{25+11}}+\frac{5}{4}\times \frac{1}{4}$

$=\sqrt{10+\sqrt{36}}+\frac{5}{16}$

$=\sqrt{10+6}+\frac{5}{16}$

$=\sqrt{16}+\frac{5}{16}$

$=4+\frac{5}{16}$

$=\frac{64+5}{16}$

$=\frac{69}{16}$

This is the required solution.