Simplify x−1xx−2+1x\dfrac{x-\frac{1}{x}}{x-2+\frac{1}{x}}. Answer ___ Make h the subject of the formula s=2h2−rs = \sqrt{2h^2 - r}. Answer ___. Given that 2ab=3\dfrac{2a}{b} = 3, a>0a > 0 and b>0b > 0, evaluate a2−54b2b\dfrac{\sqrt{a^2 - \tfrac{5}{4}b^2}}{b}. Answer ...

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Worked solution

1(a)

1. Clear the fractions in the numerator and denominator

x−1xx−2+1x=x(x−1x)x(x−2+1x)\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

=x2−1x2−2x+1=\dfrac{x^2-1}{x^2-2x+1}

3. Factorise and cancel the common factor

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

Answer

x+1x\dfrac{x+1}{x}

1(b)

1. Remove the square root by squaring both sides

s=2h2−rs=\sqrt{2h^2-r}

s2=2h2−rs^2=2h^2-r

2. Rearrange to isolate $h$

s2+r=2h2s^2+r=2h^2

h2=s2+r2h^2=\dfrac{s^2+r}{2}

h=±s2+r2h=\pm\sqrt{\dfrac{s^2+r}{2}}

Answer

h=±s2+r2h=\pm\sqrt{\dfrac{s^2+r}{2}}

1(c)

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

2ab=3\frac{2a}{b}=3
2a=3b2a=3b
a=32ba=\frac{3}{2}b

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

a2−54b2b=(32b)2−54b2b\frac{\sqrt{a^2-\frac54b^2}}{b} =\frac{\sqrt{\left(\frac32b\right)^2-\frac54b^2}}{b}

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

(94)b2−54b2b=44b2b\frac{\sqrt{\left(\frac94\right)b^2-\frac54b^2}}{b} =\frac{\sqrt{\frac44b^2}}{b}

=b2b=\frac{\sqrt{b^2}}{b}
Since b>0b>0, b2=b\sqrt{b^2}=b.
bb=1\frac{b}{b}=1

Answer

11

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