Given: \( \frac{4^{x-1}+4^{x+1}}{17.12^{x}} \) \( 2.2 .1 \quad \) Simplify the expression fully. \( 2.2 .2 \quad \) If \( 3^{-x}=4 t \), express \( \frac{4^{x-1}+4^{x+1}}{17.12^{x}} \) in terms of \( t \).

To simplify the expression \( \frac{4^{x-1}+4^{x+1}}{17.12^{x}} \), we start by rewriting \( 4^{x-1} \) and \( 4^{x+1} \). We have: \[ 4^{x-1} = \frac{4^x}{4} \quad \text{and} \quad 4^{x+1} = 4 \cdot 4^x. \] Thus, \[ 4^{x-1} + 4^{x+1} = \frac{4^x}{4} + 4 \cdot 4^x = \frac{1}{4}4^x + 4 \cdot 4^x = \frac{1 + 16}{4}4^x = \frac{17}{4}4^x. \] Now substituting this back into the original expression gives: \[ \frac{\frac{17}{4}4^x}{17.12^x} = \frac{17}{4 \cdot 17 \cdot 12^x}4^x = \frac{4^x}{4 \cdot 12^x} = \frac{4^x}{48^x} = \left(\frac{4}{48}\right)^{x} = \left(\frac{1}{12}\right)^{x}. \] Thus, the simplified expression is \( \left(\frac{1}{12}\right)^x \). Now, for the second part, we have \( 3^{-x}=4 t \), which can be rearranged to express \( 3^{x} = \frac{1}{4t} \). Using this, we need to relate it to our simplified form \( \left(\frac{1}{12}\right)^x \). Notice that: \[ 12 = 3 \times 4, \] So, \[ \left(\frac{1}{12}\right)^x = \left(\frac{1}{3 \cdot 4}\right)^x = (3^{-1} \cdot 4^{-1})^x = 3^{-x} \cdot 4^{-x}. \] From \( 3^{-x} = 4t \), we substitute to get: \[ \left(\frac{1}{12}\right)^x = (4t) \cdot 4^{-x} = 4t \cdot \left(\frac{1}{4}\right)^x = 4t \cdot \frac{1}{4^x}. \] Thus, if \( 3^{-x} = 4t \), we express \( 4^{-x} \) as \( (4^{-1})^x \). Since \( 4^{-x} = \left(\frac{1}{4}\right)^x\), we illustrate that: \[ \frac{4^{x-1}+4^{x+1}}{17.12^{x}} = \frac{4t}{4^x}. \] Therefore, the expression in terms of \( t \) is \( 4t \cdot \left(\frac{1}{4}\right)^{x} \).