If a=64 and b=289, then the value of (sqrt{sq | vedclass

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(A) Given: $a=64$ and $b=289.$First,calculate the square roots of $a$ and $b$:$\sqrt{a} = \sqrt{64} = 8$$\sqrt{b} = \sqrt{289} = 17$Now,substitute the

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$2^{\frac{1}{2}}$

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(A) Given: $a=64$ and $b=289.$First,calculate the square roots of $a$ and $b$:$\sqrt{a} = \sqrt{64} = 8$$\sqrt{b} = \sqrt{289} = 17$Now,substitute these values into the expression $(\sqrt{\sqrt{a}+\sqrt{b}}-\sqrt{\sqrt{b}-\sqrt{a}})^{\frac{1}{2}}$:$= (\sqrt{8+17} - \sqrt{17-8})^{\frac{1}{2}}$$= (\sqrt{25} - \sqrt{9})^{\frac{1}{2}}$$= (5 - 3)^{\frac{1}{2}}$$= (2)^{\frac{1}{2}}$

If $a=64$ and $b=289,$ then the value of $(\sqrt{\sqrt{a}+\sqrt{b}}-\sqrt{\sqrt{b}-\sqrt{a}})^{\frac{1}{2}}$ is

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