JIPMAT 2025Algebra > Hard1112224446\sqrt{6}6✅ Correct Option: 1Related questions:JIPMAT 2021If 2x=3y=6−z2^{x} = 3^{y} = 6^{-z}2x=3y=6−z, then (1x+1y+1z)(\frac{1}{x} + \frac{1}{y} + \frac{1}{z})(x1+y1+z1) is equal toJIPMAT 2024Simplify : 0.06254+0.0083+0.09−162⋅5×3253\dfrac{\sqrt[4]{0.0625}+\sqrt[3]{0.008}+\sqrt{0.09}-1}{\sqrt[3]{62 \cdot 5 \times \sqrt[5]{32}}}362⋅5×53240.0625+30.008+0.09−1JIPMAT 20225+38−215+11+2306−5\frac{\sqrt{5}+\sqrt{3}}{\sqrt{8-2 \sqrt{15}}}+\frac{\sqrt{11+2 \sqrt{30}}}{\sqrt{6}-\sqrt{5}}8−2155+3+6−511+230