\(\sqrt{6}\) = 2.4517, then the value of \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) correct to three places of decimal will be
\(\sqrt{1\frac{64}{225}}= 1 + \frac{?}{15}\)
\(\sqrt{300003 \times 300005}\) is nearly equal to
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