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konrad509

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triolana

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$\\\sqrt{\frac{\sqrt{113^2-112^2}}{13^2-11^2}}=\\ \sqrt{\frac{\sqrt{(113-112)(113+112)}}{(13-11)(13+11)}}=\\ \sqrt{\frac{\sqrt{1\cdot225}}{2\cdot24}}=\\ \sqrt{\frac{\sqrt{225}}{48}}=\\ \sqrt{\frac{15}{48}}=\\ \frac{\sqrt{15}}{4\sqrt3}=\\ \frac{\sqrt{45}}{12}=\\ \frac{3\sqrt5}{12}=\\ \frac{\sqrt5}{4}$

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