QUESTION: 1.I Simplifye $$ \begin{array}{l} 1.1^{\frac{25^{x+1} \cdot 25^{x-a}}{125^{x+1} \cdot 5^{x-2}}} \ 1.1 .2 \sqrt{32 x^{6}}+\sqrt{98 x^{6}} \end{array} $$ 1.2 Rationalise the denominator (2) $$ \frac{6}{4 \sqrt{x}}+\sqrt{x} $$ (2) 1.3 Solve for $$ x $$ $$ \begin{array}{l} 1.3 .13 x^{2}+9 x=12 \ 1.3 .22 x^{2}+3 x-7=0 \ 1.3 .3(3-x)(1+x)

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QUESTION: 1.I Simplifye $$ \begin{array}{l} 1.1^{\frac{25^{x+1} \cdot 25^{x-a}}{125^{x+1} \cdot 5^{x-2}}} \ 1.1 .2 \sqrt{32 x^{6}}+\sqrt{98 x^{6}} \end{array} $$ 1.2 Rationalise the denominator (2) $$ \frac{6}{4 \sqrt{x}}+\sqrt{x} $$ (2) 1.3 Solve for $$ x $$ $$ \begin{array}{l} 1.3 .13 x^{2}+9 x=12 \ 1.3 .22 x^{2}+3 x-7=0 \ 1.3 .3(3-x)(1+x)<0 \ 1.3 .4 \sqrt{x+18}+2=x \end{array} $$ (round off your answer to TWO decimal places) (3) (J) (3) 1.4 Solve for $$ x $$ and $$ y $$ $$ \begin{array}{l} 2 x^{2}+y(4 x-1)=15 \ x+y=3 \end{array} $$

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Let's solve the problems step by step. ### 1.1 Simplification #### 1.1.1 We need to simplify the expression: $$ 1^{\frac{25^{x+1} \cdot 25^{x-a}}{125^{x+1} \cdot 5^{x-2}}} $$ First, we can rewrite the bases: - $$ 25 = 5^2 $$ - $$ 125 = 5^3 $$ Now substituting these into the expression: $$ 1^{\frac{(5^2)^{x+1} \cdot (5^2)^{x-a}}{(5^3)^{x+1} \cdot 5^{x-2}}} $$ This simplifies to: $$ 1^{\frac{5^{2(x+1)} \cdot 5^{2(x-a)}}{5^{3(x+1)} \cdot 5^{x-2}}} $$ Combining the exponents in the numerator and denominator: $$ 1^{\frac{5^{2x + 2 + 2x - 2a}}{5^{3x + 3 + x - 2}}} = 1^{\frac{5^{4x - 2a}}{5^{4x + 1}}} $$ This simplifies to: $$ 1^{5^{4x - 2a - 4x - 1}} = 1^{5^{-2a - 1}} = 1 $$ So, the result is: $$ \text{Result: } 1 $$ #### 1.1.2 Now we simplify: $$ \sqrt{32 x^{6}}+\sqrt{98 x^{6}} $$ Breaking it down: $$ \sqrt{32} \cdot \sqrt{x^6} + \sqrt{98} \cdot \sqrt{x^6} $$ Calculating the square roots: $$ \sqrt{32} = 4\sqrt{2}, \quad \sqrt{98} = 7\sqrt{2}, \quad \sqrt{x^6} = x^3 $$ Thus, we have: $$ 4\sqrt{2} x^3 + 7\sqrt{2} x^3 = (4 + 7)\sqrt{2} x^3 = 11\sqrt{2} x^3 $$ So, the result is: $$ \text{Result: } 11\sqrt{2} x^3 $$ ### 1.2 Rationalise the Denominator We need to rationalise: $$ \frac{6}{4 \sqrt{x}}+\sqrt{x} $$ First, we can combine the terms: $$ \frac{6}{4\sqrt{x}} + \sqrt{x} = \frac{6 + 4x}{4\sqrt{x}} $$ Now, to rationalise the denominator, we multiply the numerator and denominator by $$\sqrt{x}$$: $$ \frac{(6 + 4x)\sqrt{x}}{4x} $$ So, the result is: $$ \text{Result: } \frac{(6 + 4x)\sqrt{x}}{4x} $$ ### 1.3 Solve for $$ x $$ #### 1.3.1 Solve: $$ 13x^{2}+9x=12 $$ Rearranging gives: $$ 13x^{2}+9x-12=0 $$ Using the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 13 \cdot (-12)}}{2 \cdot 13} $$ Calculating the discriminant: $$ 9^2 - 4 \cdot 13 \cdot (-12) = 81 + 624 = 705 $$ Now substituting back: $$ x = \frac{-9 \pm \sqrt{705}}{26} $$ Calculating $$\sqrt{705}$$: $$ \sqrt{705} \approx 26.53 $$ Thus: $$ x \approx \frac{-9 \pm 26.53}{26} $$ Calculating the two possible values: 1. $$ x_1 \approx \frac{17.53}{26} \approx 0.67 $$ 2. $$ x_2 \approx \frac{-35.53}{26} \approx -1.37 $$ Rounding to two decimal places: $$ \text{Result: } x \approx 0.67 \text{ or } -1.37 $$ #### 1.3.2 Solve: $$ 22x^{2}+3x-7=0 $$ Using the quadratic formula: $$ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 22 \cdot (-7)}}{2 \cdot 22} $$ Calculating the discriminant: $$ 3^2 + 4 \cdot 22 \cdot 7 = 9 + 616 = 625 $$ Now substituting back: $$ x = \frac{-3 \pm 25}{44} $$ Calculating the two possible values: 1. $$ x_1 = \frac{22}{44} = 0.5 $$ 2. $$ x_2 = \frac{-28}{44} \approx -0.64 $$ Rounding to two decimal places: $$ \text{Result: } x \approx 0.5 \text{ or } -0.64 $$ #### 1.3.3 Solve: $$ (3-x)(1+x)<0 $$ Finding the critical points: 1. $$ 3 - x = 0 \Rightarrow x = 3 $$ 2. $$ 1 + x = 0 \Rightarrow x = -1 $$ Testing intervals: - For $$ x < -1 $$: Choose $$ x = -2 $$ → $$ (3 - (-2))(1 + (-2)) = 5 \cdot (-1) < 0 $$ (valid) - For $$ -1 < x < 3 $$: Choose $$ x = 0 $$ → $$ (3 - 0)(1 + 0) = 3 \cdot 1 > 0 $$ (invalid) - For $$ x > 3 $$: Choose $$ x = 4 $$ → $$ (3 - 4)(1 + 4) = (-1) \cdot 5 < 0 $$ (valid) Thus, the solution is: $$ \text{Result: } x \in (-\infty, -1) \cup (3, \infty) $$ #### 1.3.4 Solve: 1.1.1: The expression simplifies to 1. 1.1.2: The expression simplifies to $$11\sqrt{2}x^3$$. 1.2: The denominator is rationalised to $$\frac{(6 + 4x)\sqrt{x}}{4x}$$. 1.3.1: The solutions are $$x \approx 0.67$$ and $$x \approx -1.37$$. 1.3.2: The solutions are $$x \approx 0.5$$ and $$x \approx -0.64$$. 1.3.3: The solution is $$x \in (-\infty, -1) \cup (3, \infty)$$. 1.3.4: The equation $$\sqrt{x+18} + 2 = x$$ has no real solutions. 1.4: The solutions are $$x = 2$$ and $$y = 1$$.

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