化简 $\dfrac{x-\frac{1}{x}}{x-2+\frac{1}{x}}$。 答 ___ 将公式 $s = \sqrt{2h^2 - r}$ 变形为以 h 为主项的形式。 答 ___。 已知 $\dfrac{2a}{b} = 3$ 且 $a > 0, b > 0$，求 $\dfrac{\sqrt{a^2 - \tfrac{5}{4}b^2}}{b}$ 的值。 答 ...

1. Clear the fractions in the numerator and denominator

$\dfrac{x-\frac{1}{x}}{x-2+\frac{1}{x}}=\dfrac{x\left(x-\frac{1}{x}\right)}{x\left(x-2+\frac{1}{x}\right)}$

2. Expand and simplify both parts

$=\dfrac{x^2-1}{x^2-2x+1}$

3. Factorise and cancel the common factor

$=\dfrac{(x-1)(x+1)}{(x-1)^2}=\dfrac{x+1}{x-1}$

1. Remove the square root by squaring both sides

$s=\sqrt{2h^2-r}$

$s^2=2h^2-r$

2. Rearrange to isolate $h$

$s^2+r=2h^2$

$h^2=\dfrac{s^2+r}{2}$

$h=\pm\sqrt{\dfrac{s^2+r}{2}}$

1. Express $a$ in terms of $b$ from the given equation

$\frac{2a}{b}=3$
$2a=3b$
$a=\frac{3}{2}b$

2. Substitute $a=\frac{3}{2}b$ into the expression

3. Simplify the square root and divide by $b$

$=\frac{\sqrt{b^2}}{b}$
Since $b>0$, $\sqrt{b^2}=b$.
$\frac{b}{b}=1$