Question

if x=√3-1/√3+1,y=√3+1/√3-1 find x²-xy+y​

Answer 1

♣ Qᴜᴇꜱᴛɪᴏɴ :

If x = $\sf{\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}$ , y = $\sf{\dfrac{\sqrt{3}+1}{\sqrt{3}-1}}$, find x²- xy + y

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{3\left(3-\sqrt{3}\right)-1}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

x = $\sf{\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}$ , y = $\sf{\dfrac{\sqrt{3}+1}{\sqrt{3}-1}}$

x²- xy + y = $\sf{\displaystyle\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^2-\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot \frac{\sqrt{3}+1}{\sqrt{3}-1}\right)+\frac{\sqrt{3}+1}{\sqrt{3}-1}}$

Solve :

$\sf{\displaystyle\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^2-\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot \frac{\sqrt{3}+1}{\sqrt{3}-1}\right)+\frac{\sqrt{3}+1}{\sqrt{3}-1}}$

$\sf{\displaystyle=\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)^2-\frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot \frac{\sqrt{3}+1}{\sqrt{3}-1}+\frac{\sqrt{3}+1}{\sqrt{3}-1}}$

$\sf{\boxed{\boxed{\begin{array}{l}\left(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right)^{2}=\dfrac{2-\sqrt{3}}{2+\sqrt{3}} \\\\\\\dfrac{\sqrt{3}-1}{\sqrt{3}+1} \cdot \dfrac{\sqrt{3}+1}{\sqrt{3}-1}=1\end{array}}}}$

$\sf{\displaystyle=\frac{2-\sqrt{3}}{2+\sqrt{3}}-1+\frac{1+\sqrt{3}}{\sqrt{3}-1}}$

$\sf{\displaystyle=\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}+1}{\sqrt{3}-1}-1}$

$\sf{=\dfrac{6\sqrt{3}}{1+\sqrt{3}}-1}$

$\large\boxed{\sf{=3\left(3-\sqrt{3}\right)-1}}$