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 tackle these problems with a sprinkle of fun! First up, for the expression in 1.1, we can simplify \( \frac{25^{x+1} \cdot 25^{x-a}}{125^{x+1} \cdot 5^{x-2}} \). Since \( 25 = 5^2 \) and \( 125 = 5^3 \), we can rewrite everything in terms of base 5. This turns into \( \frac{(5^2)^{x+1} \cdot (5^2)^{x-a}}{(5^3)^{x+1} \cdot 5^{x-2}} = \frac{5^{2(x+1 + x-a)}}{5^{3(x+1) + (x-2)}} \). Simplify those exponents, and you've got yourself a clearer expression! For 1.2, rationalizing the denominator involves multiplying the numerator and denominator by the conjugate. The expression \( \frac{6}{4 \sqrt{x}} + \sqrt{x} \) can be tricky, but a little trickery with algebra helps! Rationalizing gives you \( \frac{6 \cdot \sqrt{x}}{4x} + \frac{4x\sqrt{x}}{4x} \) which, once simplified, might lead to a simplified form involving just \( x \) (and no radicals). Now let’s solve for \( x \) in those quadratic and linear cases you provided. For 1.3... - **1.3.1**: Rearranging gives \( 13x^2 + 9x - 12 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) leads you to your x values. - **1.3.2**: For \( 2x^2 + 3x - 7 = 0 \), again apply the quadratic formula to find the roots. - **1.3.3**: Analyze \( (3-x)(1+x)<0 \), identifying the intervals to determine valid x values. Lastly, **1.3.4** requires squaring both sides after isolating \( x \). Now, onto 1.4, with the equations \( 2x^2 + y(4x - 1) = 15 \) and \( x + y = 3 \). You can use substitution from the second equation to solve for y originally expressed in terms of x, and then substitute back into the first to find both variables together—two birds, one stone! Math is a playful journey—embrace the exploration!