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If x = 16, then find the value of the ex...

If `x = 16`, then find the value of the expression `(x-1)/( x^(3//4) + x^(1//2) ) . ( x^(1//2) + x^(1//4) )/( x^(1//2) +1) . x^(1//4)`.

A

9

B

3

C

27

D

`3^0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the expression \((x-1)/( x^{3/4} + x^{1/2} ) \cdot ( x^{1/2} + x^{1/4})/( x^{1/2} +1) \cdot x^{1/4}\) for \(x = 16\), we will follow these steps: ### Step 1: Substitute the value of \(x\) We start by substituting \(x = 16\) into the expression. \[ \text{Expression} = \frac{(16-1)}{(16^{3/4} + 16^{1/2})} \cdot \frac{(16^{1/2} + 16^{1/4})}{(16^{1/2} + 1)} \cdot 16^{1/4} \] ### Step 2: Simplify the numerator Calculate \(16 - 1\): \[ 16 - 1 = 15 \] ### Step 3: Calculate the powers of \(16\) Now, calculate \(16^{3/4}\), \(16^{1/2}\), and \(16^{1/4}\): - \(16^{1/2} = \sqrt{16} = 4\) - \(16^{1/4} = \sqrt[4]{16} = 2\) - \(16^{3/4} = (16^{1/4})^3 = 2^3 = 8\) ### Step 4: Substitute these values back into the expression Now substitute these values back into the expression: \[ \text{Expression} = \frac{15}{(8 + 4)} \cdot \frac{(4 + 2)}{(4 + 1)} \cdot 2 \] ### Step 5: Simplify the denominator Calculate \(8 + 4\) and \(4 + 1\): \[ 8 + 4 = 12 \] \[ 4 + 1 = 5 \] ### Step 6: Substitute and simplify further Now substitute these values back into the expression: \[ \text{Expression} = \frac{15}{12} \cdot \frac{6}{5} \cdot 2 \] ### Step 7: Simplify \(\frac{15}{12}\) Simplify \(\frac{15}{12}\): \[ \frac{15}{12} = \frac{5}{4} \] ### Step 8: Substitute and simplify Now substitute this back into the expression: \[ \text{Expression} = \frac{5}{4} \cdot \frac{6}{5} \cdot 2 \] ### Step 9: Cancel out the \(5\) The \(5\) in the numerator and denominator cancels out: \[ \text{Expression} = \frac{6}{4} \cdot 2 \] ### Step 10: Simplify \(\frac{6}{4}\) Simplify \(\frac{6}{4}\): \[ \frac{6}{4} = \frac{3}{2} \] ### Step 11: Final multiplication Now multiply \(\frac{3}{2} \cdot 2\): \[ \text{Expression} = 3 \] ### Final Answer Thus, the value of the expression when \(x = 16\) is: \[ \boxed{3} \]
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