1) 1/(Square Root of x) - 1 / x - 1
So I multiply both top and bottom by 1/(Square Root of x) + 1 , right? But I don't know what that equals, I just don't know how to multiply them...
2) x(Square Root of x) - 8 / x - 4
Same thing...maybe if I get help on 1 I can get this one.
More questions to come I'm sure...
If you typed more intelligibly, it would be easier t answer you and you would have probably gotten an answer by now.Originally Posted by Dickson
what you typed can be interpreted several ways, please use parentheses to clarify
i believe the first question is $\displaystyle \frac {\frac 1{\sqrt{x}} - 1}{x - 1}$. personally, i'd combine the fractions first, but multiplying by $\displaystyle \frac { \frac 1{\sqrt{x}} + 1}{ \frac 1{\sqrt{x}} + 1}$ is a valid option.
note that the whole point of multiplying by the conjugate is to obtain the difference of two squares. hence the numerator will change from the form $\displaystyle (x - y)(x + y)$ to the form $\displaystyle x^2 - y^2$, and we get
$\displaystyle \frac {\frac 1x - 1}{(x - 1) \left( \frac 1{\sqrt{x}} + 1\right)}$
now simplify
for the second problem, note that you have $\displaystyle \frac {(x^{1/2})^3 - 8}{x - 4}$
which should remind you of the difference of two cubes formula, which should help you know how to proceed.
Note that the key is knowing the formula: $\displaystyle (a+b)(a-b)=a^2-b^2$
$\displaystyle \frac{\frac{1}{\sqrt{x}}-1}{x-1} \times \frac{\frac{1}{\sqrt{x}}+1}{\frac{1}{\sqrt{x}}+1}= \frac{\frac{1}{x}-1}{(x-1)(\frac{1}{\sqrt{x}}+1)}=...$
See if you can take it from here.
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