Question

pls solve the question in the image​

Answer 1

Answer :

1

___________

As per the provided information in the given question, we have :

• $\rm {x = \dfrac{1}{7+4\sqrt{3} } }$

• $\rm {y= \dfrac{1}{7-4\sqrt{3} } }$

We are asked to calculate the value of ,

$\longmapsto \bf { \dfrac{1}{x +1} + \dfrac{1}{y + 1} }$

Here, let us first rationalise the denominator of the the value of x and y.

In order to rationalise the denominator of any fraction, we multiply the rationalising factor of the denominator with the numerator and the denominator of the fraction.

Rationalising the denominator of :

$\longmapsto \rm {x = \dfrac{1}{7+4\sqrt{3} } }$

Rationalising factor of (a + b) is (a - b). Here, denominator is in the form of (a + b), so its rationalising factor is (7 - 4√3). Multiplying (7 - 4√3) with both the numerator and the denominator of the fraction.

$\longmapsto \rm {x = \dfrac{1}{7+4\sqrt{3} } \times \dfrac{7-4\sqrt{3}}{7-4\sqrt{3}} }$

Rearranging the terms.

$\longmapsto \rm {x = \dfrac{1(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3}) } }$

Performing multiplication and using the identity,

$\rm {(a+b)(a-b) = a^2 - b^2}$

$\longmapsto \rm {x = \dfrac{7-4\sqrt{3}}{(7)^2-(4\sqrt{3})^2 } }$

Writing the squares of the numbers in the denominator.

$\longmapsto \rm {x = \dfrac{7-4\sqrt{3}}{49-48 } }$

Performing subtraction.

$\longmapsto \rm {x = \dfrac{7-4\sqrt{3}}{1 } }$

We can write x as,

$\longmapsto \bf{ x = 7-4\sqrt{3}}$

Rationalising the denominator of :

$\longmapsto \rm {y = \dfrac{1}{7-4\sqrt{3} } }$

Rationalising factor of (a - b) is (a + b). Here, denominator is in the form of (a - b), so its rationalising factor is (7 + 4√3). Multiplying (7 + 4√3) with both the numerator and the denominator of the fraction.

$\longmapsto \rm {y = \dfrac{1}{7-4\sqrt{3} } \times \dfrac{7+4\sqrt{3}}{7+4\sqrt{3}} }$

Rearranging the terms.

$\longmapsto \rm {y = \dfrac{1(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3}) } }$

Performing multiplication and using the identity,

$\rm {(a+b)(a-b) = a^2 - b^2}$

$\longmapsto \rm {y = \dfrac{7+4\sqrt{3}}{(7)^2-(4\sqrt{3})^2 } }$

Writing the squares of the numbers in the denominator.

$\longmapsto \rm {y = \dfrac{7+4\sqrt{3}}{49-48 } }$

Performing subtraction.

$\longmapsto \rm {y = \dfrac{7+4\sqrt{3}}{1 } }$

We can write x as,

$\longmapsto \bf{ y = 7+4\sqrt{3}}$

Now, finding the required answer :

$\longmapsto \bf { \dfrac{1}{x +1} + \dfrac{1}{y + 1} }$

Substituting the value of x and y.

$\longmapsto \rm { \dfrac{1}{(7-4\sqrt{3})+1} + \dfrac{1}{(7+4\sqrt{3} )+ 1} }$

Removing the brackets.

$\longmapsto \rm { \dfrac{1}{7-4\sqrt{3}+1} + \dfrac{1}{7+4\sqrt{3} + 1} }$

Performing addition in the denominator.

$\longmapsto \rm { \dfrac{1}{8-4\sqrt{3}} + \dfrac{1}{8+4\sqrt{3}} }$

Now, taking the L.C.M and simplifying further.

$\longmapsto \rm { \dfrac{(8+4\sqrt{3})+(8 -4\sqrt{3})}{(8-4\sqrt{3})(8+4\sqrt{3})} }$

Removing the brackets in the numerator.

$\longmapsto \rm { \dfrac{8+4\sqrt{3}+8 -4\sqrt{3}}{(8-4\sqrt{3})(8+4\sqrt{3})} }$

Performing addition and subtraction and using the identity,

$\rm {(a+b)(a-b) = a^2 - b^2}$

$\longmapsto \rm { \dfrac{16\cancel{+4\sqrt{3} -4\sqrt{3}} }{(8)^2 - (4\sqrt{3})^2} }$

Simplifying further.

$\longmapsto \rm { \dfrac{16}{64 - 48} }$

Performing substraction in denominator.

$\longmapsto \rm { \dfrac{16}{16} }$

Dividing 16 by 16.

$\longmapsto \bf { 1 }$

∴ The required answer is 1.