1/√(5+√24) +3/√(9+√72) -4/√(8+√48)

asked by guest
on Feb 02, 2025 at 4:16 am



You asked:

Evaluate the expression: 15+24+39+72−48+48\frac{1}{\sqrt{5 + \sqrt{24}}} + \frac{3}{\sqrt{9 + \sqrt{72}}} - \frac{4}{\sqrt{8 + \sqrt{48}}}

MathBot Answer:

15+24+39+72−48+48=−443+8+126+5+362+9\frac{1}{\sqrt{5 + \sqrt{24}}} + \frac{3}{\sqrt{9 + \sqrt{72}}} - \frac{4}{\sqrt{8 + \sqrt{48}}} = - \frac{4}{\sqrt{4 \sqrt{3} + 8}} + \frac{1}{\sqrt{2 \sqrt{6} + 5}} + \frac{3}{\sqrt{6 \sqrt{2} + 9}}

15+24+39+72−48+48=15+24+3⋅19+72−4⋅18+48=15+2⋅6+3⋅19+72−4⋅18+48=15+2⋅6+3⋅19+6⋅2−4⋅18+48=15+2⋅6+3⋅19+6⋅2−4⋅18+4⋅3=(15+2⋅6+3⋅19+6⋅2)−4⋅18+4⋅3=15+2⋅6−4⋅18+4⋅3+3⋅19+6⋅2\begin{aligned}\frac{1}{\sqrt{5 + \sqrt{24}}} + \frac{3}{\sqrt{9 + \sqrt{72}}} - \frac{4}{\sqrt{8 + \sqrt{48}}}& = \frac{1}{\sqrt{5 + \sqrt{24}}} + 3 \cdot \frac{1}{\sqrt{9 + \sqrt{72}}} - 4 \cdot \frac{1}{\sqrt{8 + \sqrt{48}}}\\& = \frac{1}{\sqrt{5 + 2 \cdot \sqrt{6}}} + 3 \cdot \frac{1}{\sqrt{9 + \sqrt{72}}} - 4 \cdot \frac{1}{\sqrt{8 + \sqrt{48}}}\\& = \frac{1}{\sqrt{5 + 2 \cdot \sqrt{6}}} + 3 \cdot \frac{1}{\sqrt{9 + 6 \cdot \sqrt{2}}} - 4 \cdot \frac{1}{\sqrt{8 + \sqrt{48}}}\\& = \frac{1}{\sqrt{5 + 2 \cdot \sqrt{6}}} + 3 \cdot \frac{1}{\sqrt{9 + 6 \cdot \sqrt{2}}} - 4 \cdot \frac{1}{\sqrt{8 + 4 \cdot \sqrt{3}}}\\& = \left(\frac{1}{\sqrt{5 + 2 \cdot \sqrt{6}}} + 3 \cdot \frac{1}{\sqrt{9 + 6 \cdot \sqrt{2}}}\right) - 4 \cdot \frac{1}{\sqrt{8 + 4 \cdot \sqrt{3}}}\\& = \frac{1}{\sqrt{5 + 2 \cdot \sqrt{6}}} - 4 \cdot \frac{1}{\sqrt{8 + 4 \cdot \sqrt{3}}} + 3 \cdot \frac{1}{\sqrt{9 + 6 \cdot \sqrt{2}}}\end{aligned}

+ Submit Question