417−214312+117÷12+12+15−15−−−−−−−−−−−−−−−−−−−⎷$$\sqrt{\frac{4\frac{1}{7}-2\frac{1}{4}}{3\frac{1}{2}+1\frac{1}{7}}÷\frac{1}{2+\frac{1}{2+\frac{1}{5-\frac{1}{5}}}}}$$

$$\sqrt {\frac{{4\frac{1}{7} - 2\frac{1}{4}}}{{3\frac{1}{2} + 1\frac{1}{7}}} \div \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{5 - \frac{1}{5}}}}}}}}$$     is equal to = ?

Explanation:-

$$\eqalign{ & {\text{Take first part }} \cr & \frac{{4\frac{1}{7} - 2\frac{1}{4}}}{{3\frac{1}{2} + 1\frac{1}{7}}} \cr & = \frac{{\frac{{29}}{7} - \frac{9}{4}}}{{\frac{7}{2} + \frac{8}{7}}} \cr & = \frac{{\frac{{116 - 63}}{{28}}}}{{\frac{{49 + 16}}{{14}}}} \cr & = \frac{{53}}{{28}} \times \frac{{14}}{{65}} \cr & = \frac{{53}}{{130}} \cr & {\text{The second part}} \cr & \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{5 - \frac{1}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{\frac{{25 - 1}}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{5}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{\frac{{53}}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{{24}}{{53}}}} \cr & = \frac{1}{{\frac{{106 + 24}}{{53}}}} \cr & = \frac{{53}}{{130}} \cr & {\text{According to question}} \cr & \,\,\,\,\,\,\sqrt {\frac{{53}}{{130}} \div \frac{{53}}{{130}}} \cr & = \sqrt {\frac{{53}}{{130}} \times \frac{{130}}{{63}}} \cr & = \sqrt 1 \cr & = 1 \cr}$$