\frac{\sqrt{10^2-2^6}}{\sqrt{\frac{3^3-2^3}{2^2}}}

asked by guest
on Nov 25, 2024 at 6:40 pm



You asked:

Evaluate the expression: 102−2633−2322\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}}

MathBot Answer:

102−2633−2322=121919≈2.75298880644674119088651493717449\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}} = \frac{12 \sqrt{19}}{19} \approx 2.75298880644674119088651493717449

102−2633−2322=102−26⋅1(33−23)⋅122=100−26⋅1(33−23)⋅122=100−64⋅1(33−23)⋅122=36⋅1(33−23)⋅122=6⋅1(33−23)⋅122=6⋅1(27−23)⋅122=6⋅1(27−8)⋅122=6⋅119⋅122=6⋅1194=6⋅112⋅19=1219⋅19\begin{aligned}\frac{\sqrt{{10}^{2} - {2}^{6}}}{\sqrt{\frac{{3}^{3} - {2}^{3}}{{2}^{2}}}}&=\sqrt{10^{2} - 2^{6}} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=\sqrt{100 - 2^{6}} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=\sqrt{100 - 64} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=\sqrt{36} \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=6 \cdot \frac{1}{\sqrt{\left(3^{3} - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=6 \cdot \frac{1}{\sqrt{\left(27 - 2^{3}\right) \cdot \frac{1}{2^{2}}}}\\&=6 \cdot \frac{1}{\sqrt{\left(27 - 8\right) \cdot \frac{1}{2^{2}}}}\\&=6 \cdot \frac{1}{\sqrt{19 \cdot \frac{1}{2^{2}}}}\\&=6 \cdot \frac{1}{\sqrt{\frac{19}{4}}}\\&=6 \cdot \frac{1}{\frac{1}{2} \cdot \sqrt{19}}\\&=\frac{12}{19} \cdot \sqrt{19}\end{aligned}

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