化简 (i) $\dfrac{1}{\sqrt[a]{x}}\times b + \left(\dfrac{2}{ab}\right)^{-2}$。 答 ___ (ii) $\dfrac{4x^2-36}{2x^2-20x+42}$。 答 ___

1. Rewrite the reciprocal of the $a$th root using fractional indices.

$\sqrt[a]{x}=x^{\frac{1}{a}}$

So

$\dfrac{1}{\sqrt[a]{x}}\times b=\dfrac{b}{x^{\frac{1}{a}}}=bx^{-\frac{1}{a}}$.

2. Simplify the term with the negative power.

$\left(\dfrac{2}{ab}\right)^{-2}=\left(\dfrac{ab}{2}\right)^2$

3. Square the numerator and denominator and write the simplified expression.

$\left(\dfrac{ab}{2}\right)^2=\dfrac{a^2b^2}{4}$

Hence,

$\dfrac{1}{\sqrt[a]{x}}\times b+\left(\dfrac{2}{ab}\right)^{-2}=\dfrac{b}{\sqrt[a]{x}}+\dfrac{a^2b^2}{4}$.

1. Factorise the numerator by taking out the common factor and using difference of two squares.

$4x^2-36=4(x^2-9)=4(x+3)(x-3)$

2. Factorise the denominator by taking out the common factor and factorising the quadratic.

$2x^2-20x+42=2(x^2-10x+21)=2(x-3)(x-7)$

3. Cancel the common factor and simplify the numerical coefficient.

$\frac{4x^2-36}{2x^2-20x+42}=\frac{4(x+3)(x-3)}{2(x-3)(x-7)}=\frac{2(x+3)}{x-7}$