Rationalise the denominator.
1
---------
5 - √2
Will give brainiest and loads of points.
Answers
Let's see what to do buddy...
_________________________________
To do this, we have to multiply the face and denominator of the fraction by the denominator conjunction, which is :
[tex](5 -\sqrt{2}) [/tex]
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
Reminder :
[tex](a + b)(a - b) = {a}^{2} - {b}^{2} \\ [/tex]
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
So we have :
[tex] \frac{1}{ 5-\sqrt{2} } \times \frac{ 5+\sqrt{2} }{ 5 +\sqrt{2} } = \frac{1 \times ( 5 +\sqrt{2}) }{(5- \sqrt{2} )( 5 +\sqrt{2}) } = \\ [/tex]
The other is in the photo.
_________________________________
And we're done.
Thanks for watching buddy good luck.
♥️♥️♥️♥️♥️
Answer:
⇛(5+√2)/23.
Step-by-step explanation:
Given,
1/(5-√2)
The denominator is 5-√2.
We know that
The rationalising factor of a+√b is a-√b.
Therefore, the rationalising factor of 5-√2 is 5+√2. To rationalise the denominator is 1/(5-√2), we multiply this by (5+√2)/(5+√2).
∴ 1/(5-√2)
= {1/(5-√2)}×{(5+√2)/(5+√2)}
= {1(5+√2)}/{(5)²-(√2)²} [∵(a-b)(a+b)=a²-b²]
= {1(5+√2)}/{(5*5)-(√2*2)}
= {1(5+√2)}/25 - 2
= {1(5+√2)}/23
= (5+√2)/23
Hence the denominator is rationalised.
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(
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x
⋅
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(
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