$$\frac{\sqrt{26} - \sqrt{13} - \sqrt{2}+1}{\sqrt{13} -1 }= ? $$
My attempt:
$$\frac {\sqrt{2}. \sqrt{13} - \sqrt{13}-2 +1}{2} \tag 1$$ Which equals to $$\frac {\sqrt{2}-1}{\sqrt{13} +1} \tag 2 = ...$$
Waiting for your helps.
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Typesetting tip: To obtain $\sqrt{13}$, type \sqrt{13} when you are in math mode. – N. F. Taussig Feb 19 '18 at 01:51
2 Answers
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\begin{align}\frac{\sqrt 2 \cdot \sqrt{13}-\sqrt{13}-\sqrt2+1}{\sqrt{13}-1}&=\frac{\sqrt{13}(\sqrt 2 -1)-1 (\sqrt2-1)}{\sqrt{13}-1}\\&=\frac{(\sqrt{13}-1)(\sqrt 2-1)}{\sqrt{13}-1}\\&=\sqrt2-1 \end{align}
Jaideep Khare
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@Liam I have applied the property $ab-b=(a-1)b$. Here $b=\sqrt{13}$ and $a=\sqrt 2$. – Jaideep Khare Feb 18 '18 at 22:14
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@Liam how will you split $10 \cdot (5+3)$ Can't you write it as $10 \cdot 5+ 10 \cdot 3$. We can, right? Similarly, $a(b+c)=ab+ac$. Just apply this in reverse – Jaideep Khare Feb 18 '18 at 22:19
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@Liam $\sqrt{13}(\sqrt2 -1) = \sqrt{13} \cdot \sqrt 2- \sqrt{13} \cdot 1 = \sqrt{13} \cdot \sqrt{2}-\sqrt{13}= \sqrt 2 \cdot \sqrt{13} -\sqrt {13}$. Just go reverse in each step. I hope you'll understand. – Jaideep Khare Feb 18 '18 at 22:21
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I got that, what I didn't get is $-1(\sqrt 2 - 1)$. Oh sorry I got it now!! am too grateful, thank you – Liam Feb 18 '18 at 22:22
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$$\dfrac{\sqrt 2-1}{\sqrt{13}+1}=\dfrac{\sqrt 2-1}{\sqrt{13}+1}\dfrac{\sqrt {13}-1}{\sqrt{13}-1}=\dfrac{1}{12}(\sqrt2-1)(\sqrt{13}-1)=\dfrac{1}{12}(\sqrt{26}-\sqrt{13}-\sqrt{2}+1)$$
Mostafa Ayaz
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