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Question: \[\sqrt{\mathbf{\lbrack 10}\mathbf{-}\sqrt{\mathbf{(24)}}\mathbf{-}\sqrt{\mathbf{(40)}}\mathbf{+}\sq...

[10−(24)−(40)+(60)]=\sqrt{\mathbf{\lbrack 10}\mathbf{-}\sqrt{\mathbf{(24)}}\mathbf{-}\sqrt{\mathbf{(40)}}\mathbf{+}\sqrt{\mathbf{(60)}}\mathbf{\rbrack}}\mathbf{=}

A

5+3+2\sqrt{5} + \sqrt{3} + \sqrt{2}

B

5+3−2\sqrt{5} + \sqrt{3} - \sqrt{2}

C

5−3+2\sqrt{5} - \sqrt{3} + \sqrt{2}

D

2+3−5\sqrt{2} + \sqrt{3} - \sqrt{5}

Answer

5+3−2\sqrt{5} + \sqrt{3} - \sqrt{2}

Explanation

Solution

Let 10−24−40+60=(a−b+c)210 - \sqrt{24} - \sqrt{40} + \sqrt{60} = (\sqrt{a} - \sqrt{b} + \sqrt{c})^{2}

10−24−40+60=a+b+c−2ab−2bc+2ca,a,b,c>010 - \sqrt{24} - \sqrt{40} + \sqrt{60} = a + b + c - 2\sqrt{ab} - 2\sqrt{bc} + 2\sqrt{ca},a,b,c > 0Then a+b+c=10,a + b + c = 10, ab=6ab = 6, bc=10,ca=15bc = 10,ca = 15

a2b2c2=900a^{2}b^{2}c^{2} = 900⇒ abc = 30 (≠±30)( \neq \pm 30). So a=3,b=2,c=5a = 3,b = 2,c = 5

Therefore, (10−24−40+60)=±(3+5−2)\sqrt{(10 - \sqrt{24} - \sqrt{40} + \sqrt{60})} = \pm (\sqrt{3} + \sqrt{5} - \sqrt{2})