[tex] \frac{1}{ \sqrt{7} -(- 2)} \\ rationalizing \: \: \: the \: \: \: denominator \: \: \: we \: \: \: have \\ = \frac{1}{ \sqrt{7} + 2 } \times \frac{ \sqrt{7} - 2}{ \sqrt{7} - 2} \\ = \frac{ \sqrt{7} - 2}{ {( \sqrt{7} )}^{2} - {(2)}^{2} } \\ = \frac{ \sqrt{7} - 2 }{7 - 4} \\ = \frac{ \sqrt{7} - 2}{3} [/tex]
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Given that:
1/(√7 - 2)
The denominator is (√7-2).
We know
The rationalising factor of (√a-b) is (√a+b)
Therefore, the rationalising factor of √7-2 is √7+2.
On rationalising the denominator them
⇛[1/(√7-2)]×[(√7+2)/(√7+2)]
⇛[1(√7+2)]/[(√7-2)(√7+2)]
Since, (a-b)(a+b) = a²-b²
Where, a = √7 and b = √2.
⇛[1(√7+2)]/[(√7)²-(2)²]
⇛[1(√7+2)]/[(√7*7)-(2*2)]
⇛[1(√7+2)]/[7-4]
⇛[1(√7+2)]/3
⇛(√7+2)/3
Hence, the denominator is rationalised.
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