QUESTION 2 2.1 Simplify the following without using a calcu negative exponent and irrational exponents. 2.1.1 \( \frac{3^{2 n+2}-9^{n-1}}{2.3^{2 n+1}} \) 2.1.2 \( \frac{(3-\sqrt{3})^{2}}{\sqrt{3} \cdot \sqrt{6}} \) Solve for \( x \), withount usimg a callcullator. 2.2 2.2.1 \( \quad \sqrt[x]{32}=128 \) \( 2.2 .2 \quad x^{\frac{1}{2}}-3 x^{\frac{1}{4}}-10=0 \)

To simplify \( \frac{3^{2n+2} - 9^{n-1}}{2 \cdot 3^{2n+1}} \), first notice that \( 9 \) can be rewritten as \( 3^2 \). Thus, \( 9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2} \). This means we can express the numerator as \( 3^{2n+2} - 3^{2n-2} \). Factoring out \( 3^{2n-2} \), we get \( 3^{2n-2}(3^4 - 1) \). Now, simplify the expression further by substituting this back into the fraction. To simplify \( \frac{(3 - \sqrt{3})^2}{\sqrt{3} \cdot \sqrt{6}} \), start by expanding the numerator: \( (3 - \sqrt{3})^2 = 9 - 6\sqrt{3} + 3 = 12 - 6\sqrt{3} \). The denominator can be rewritten as \( \sqrt{18} \) (since \( \sqrt{6} = \sqrt{2 \cdot 3} \)). Continue simplifying to express the entire ratio in its simplest form. For the equation \( \sqrt[x]{32}=128 \), rewrite it as \( 32^{1/x} = 128 \). Knowing that \( 32 = 2^5 \) (so \( 32^{1/x} = 2^{5/x} \)) and \( 128 = 2^7 \), you can set up the equation \( \frac{5}{x} = 7 \), leading to \( x = \frac{5}{7} \). For the equation \( x^{1/2} - 3x^{1/4} - 10 = 0 \), substitute \( y = x^{1/4} \), giving you the equation \( y^2 - 3y - 10 = 0 \). This quadratic equation can be solved using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). With \( a = 1, b = -3, c = -10 \), find the two values of \( y \) and then back-calculate to find \( x \).