Question

If $ x=\frac{\left[ \begin{align} & 729+6(2)(243)+15(4)(81)+20(8)(27) \\ & +15(16)(9)+6(32)3+64 \\ \end{align} \right]}{1+4(4)+6(16)+4(64)+256} $ then $ \sqrt{x}-\frac{1}{\sqrt{x}} $ is equal to:

• A. 0.2
• B. 4.8
• C. 1.02
• D. 5.2
• E. 25

Answer 1

Answer: (B)

$ x=\frac{\left[ \begin{align} & 729+6(2)(243)+15(4)(81)+20 \\ & \times (8)(27)+15(16)(9)+6(32)(3)+64 \\ \end{align} \right]}{1+4(4)+6(16)+4(64)+256} $ $ =\frac{\left[ \begin{align} & ^{6}{{C}_{0}}{{(3)}^{6}}{{+}^{6}}{{C}_{1}}{{3}^{5}}.2{{+}^{6}}{{C}_{2}}{{3}^{4}}{{.2}^{2}}{{+}^{6}}{{C}_{3}} \\ & \times {{3}^{3}}{{2}^{2}}{{+}^{6}}{{C}_{4}}{{.3}^{2}}{{.2}^{4}}{{+}^{6}}{{C}_{5}}{{3.2}^{5}}{{+}^{6}}{{C}_{6}}{{2}^{6}} \\ \end{align} \right]}{\left[ \begin{align} & ^{4}{{C}_{0}}{{1}^{4}}{{+}^{4}}{{C}_{1}}4{{+}^{4}}{{C}_{2}}{{4}^{2}}{{+}^{4}}{{C}_{3}}{{4}^{3}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{4}}{{C}_{4}}{{4}^{4}} \\ \end{align} \right]} $ $ \Rightarrow $ $ x=\frac{{{(3+2)}^{6}}}{{{(1+4)}^{4}}}={{5}^{2}} $ $ \therefore $ $ \sqrt{x}=5 $ $ \therefore $ $ \sqrt{x}-\frac{1}{\sqrt{x}}=5-\frac{1}{5} $ $ =\frac{24}{5}=4.8 $