Question

Rationalise the denominator:-
1/(√7+√5-√3)​

Answer 1

Answer:

NOTE:

• The question could be wrong... I had got a BIZZARE answer at the end!
• Please mark me as the branliest... this question took so long and i had put in a lot of effort!

$\frac{1}{\sqrt{7}+\sqrt{5}-\sqrt{3}}$.

Rationalising factor of this would be $(\sqrt{7} +\sqrt{5} )+\sqrt{3}$.

In case you didn't know rationalising factor, the factor of multiplication by which rationalization is done is called a rationalizing factor. Like if √2 is multiplied with √2, it will 2, which is a rational number, so √2 is rationalizing factor of √2.

Common examples:

(a + √x) and (a + √x) are rationalizing factors of each other.

(a + b√x) and (a - b√x) are rationalizing factors of each other.

√x + √y and √x - √y are rationalizing factors of each other.

Now, let's get back with the sum!

we multiply the rationalising factor on both numerator and denominator.

$\frac{1}{\sqrt{7}+\sqrt{5}-\sqrt{3}}*\frac{(\sqrt{7} +\sqrt{5} )+\sqrt{3}}{(\sqrt{7} +\sqrt{5} )+\sqrt{3}}$

Well, now multiply... sounds simple but this is the tough part!

$\frac{(\sqrt{7} +\sqrt{5} )+\sqrt{3}}{\sqrt{7}+\sqrt{5}-\sqrt{3}*(\sqrt{7} +\sqrt{5} )+\sqrt{3}}$

take (√7+√5) as x, take √3 as y.............................(1)

Thus, replacing the values in the above expression, we get:

$\frac{\sqrt{7}+\sqrt{5}+\sqrt{3} }{(x+y)*(x-y)}$

We know that (x-y)(x+y)=x²-y²

So, we can rewrite the expression as:

$\frac{\sqrt{7}+\sqrt{5}-\sqrt{3}}{x^{2}-y^{2} }$

we know that (√7+√5) is x, √3 is y.........(from 1)

Then just substitute these numbers in the places of x and y.

$\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{(\sqrt{7}+\sqrt{2} )^{2}-\sqrt{3} ^{2} }$.

√3², √ and ² will get cancelled. thus, √3²=3.

And, (√7+√2)² is in the form (a+b)²=a²+b²+2ab.

thus, (√7+√2)²=√7²+√2²+2×√7×√2

(√7+√2)² =9+√14.

Thus, √3²=3,  (√7+√2)²=9+√14.

Thus, substitute.

$\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{9+\sqrt{14} -3}$

$\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{6+\sqrt{14} }$.

Again, we multiply the denominator and numerator with the rationalising factor.

here, the rationalising factor is 6-√14. Thus,

$\frac{\sqrt{7}+\sqrt{5}+\sqrt{3}}{6+\sqrt{14} }*\frac{6-\sqrt{14} }{6-\sqrt{14} }$

Here, multiplication of (√7+√5+√3)(6-√14) is very painful, but do it for the marks.

(√7+√5+√3)(6−√14)

=

Thus,

$\frac{-7\sqrt{2}+6\sqrt{3}-\sqrt{42}+6\sqrt{5}+6\sqrt{7}-\sqrt{70}}{6^{2}-\sqrt{14} ^{2} }$.

=$\frac{-7\sqrt{2}+6\sqrt{3}-\sqrt{42}+6\sqrt{5}+6\sqrt{7}-\sqrt{70}}{50}$.

If the question is wrong... IDC!

Hope this answer is instructive...

HOPE THIS HELPS :D

Answer 2

Step-by-step explanation:

$1/(√7+√3-√2)</p><p></p><p>multiply numerator and denominator by (√7+√3+√2)</p><p></p><p>=(√7+√3+√2)/[(√7+√3)-(√2)][(√7+√3)+(√2)]</p><p></p><p>=(√7+√3+√2)/(√7+√3)²-(√2)²</p><p></p><p>=(√7+√3+√2)/(√7)²+(√3)²+2(√7)(√3)-2</p><p></p><p>=(√7+√3+√2)/7+3+2(√21)-2</p><p></p><p>=(√7+√3+√2)/8+2(√21)</p><p></p><p>multiply numerator and denominator by 8–2(√21)</p><p></p><p>(√7+√3+√2)[8–2(√21)]/[8+2(√21)][8–2(√21)]</p><p></p><p>=(√7+√3+√2)[8–2(√21)]/[8]²-[2(√21)]²</p><p></p><p>=(√7+√3+√2)[8–2(√21)]/64–4*21</p><p></p><p>=(√7+√3+√2)[8–2(√21)]/64–84</p><p></p><p>=-(√7+√3+√2)[8–2(√21)]/20</p><p></p><p>you can solve it further</p><p></p><p>=-(√7+√3+√2)2[4–(√21)]/20</p><p></p><p>=-(√7+√3+√2)(4–√21)/10</p><p></p><p>or further</p><p></p><p>=-4√7–4√3–4√2+√21*7+√21*3+√21*2/10</p><p></p><p>=-4√7–4√3–4√2+7√3+3√7+√42/10</p><p></p><p>solving similar terms</p><p></p><p>=[-√7+3√3–4√2+√42]/10</p><p></p><p>$

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