If \(6^x - 6^{x-1} = 30\), then the value of \(\dfrac{2x-1}{2x+3}\) is:
\(\tfrac{3}{7}\)
\(6^x - 6^{x-1} = 30\) can be written as \(6^{x-1}(6-1) = 30\), so \(6^{x-1}\times5=30\).
This gives \(6^{x-1}=6\), so \(x-1=1\), meaning \(x=2\).
Substituting into the required expression: \(\dfrac{2x-1}{2x+3} = \dfrac{2(2)-1}{2(2)+3} = \dfrac{3}{7}\).
Hence, the value of the expression is \(\tfrac{3}{7}\).
The value of (0.3) [{(200 - 146)/(3 × 3 × 3)} - 3] is:
The expression \(\frac{{15\left( {\sqrt {10} + \sqrt 5 } \right)}}{{\sqrt {10\;} + \sqrt {20} + \sqrt {40} - \sqrt 5 - \sqrt {80} }}\) is equal to:
Let \(x = \left( {\frac{{√ {1875} }}{{√ {3888} }} \div \frac{{√ {1200} }}{{\sqrt 768}}} \right) \times \frac{{√ {175} }}{{√ {1792} }}\) . Then √x is equal to:
If \(x = \sqrt {-\sqrt 3 + \sqrt {3 + 8\sqrt {7 + 4\sqrt 3}}}\) where x > 0, then the value of x is equal to:
What is the value of \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}−\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}−\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}\) ?