Answered

Join the IDNLearn.com community and start exploring a world of knowledge today. Discover in-depth answers to your questions from our community of experienced professionals.

57. सरल गर (Simplify): [tex]\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}[/tex]

a. 14
b. 5
c. 3
d. 6

Sagot :

GinnyAnsweridn
To simplify the given expression [tex]\(\frac{1}{\sqrt{5}-2} - \frac{1}{\sqrt{5}+2}\)[/tex], we follow these steps:

### Step 1: Rationalizing the Denominators

To rationalize the denominators, we multiply the numerator and denominator by the conjugate of the denominator.

#### Rationalizing [tex]\(\frac{1}{\sqrt{5}-2}\)[/tex]

The conjugate of [tex]\(\sqrt{5} - 2\)[/tex] is [tex]\(\sqrt{5} + 2\)[/tex]. So,
[tex]\[ \frac{1}{\sqrt{5} - 2} \cdot \frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} = \frac{\sqrt{5} + 2}{5 - 4} = \frac{\sqrt{5} + 2}{1} = \sqrt{5} + 2 \][/tex]

#### Rationalizing [tex]\(\frac{1}{\sqrt{5}+2}\)[/tex]

The conjugate of [tex]\(\sqrt{5} + 2\)[/tex] is [tex]\(\sqrt{5} - 2\)[/tex]. So,
[tex]\[ \frac{1}{\sqrt{5} + 2} \cdot \frac{\sqrt{5} - 2}{\sqrt{5} - 2} = \frac{\sqrt{5} - 2}{(\sqrt{5})^2 - (2)^2} = \frac{\sqrt{5} - 2}{5 - 4} = \frac{\sqrt{5} - 2}{1} = \sqrt{5} - 2 \][/tex]

### Step 2: Subtracting the Rationalized Expressions

Now, we subtract the simplified expressions:
[tex]\[ (\sqrt{5} + 2) - (\sqrt{5} - 2) \][/tex]

Distribute the subtraction:
[tex]\[ \sqrt{5} + 2 - \sqrt{5} + 2 = 2 + 2 = 4 \][/tex]

So, the simplified form of the given expression is [tex]\(4\)[/tex].

### Conclusion

The correct answer is not directly provided in the options, which suggests a need for reassessment of the problem. Considering typical examination contexts, this step-by-step calculation's conclusion of [tex]\(\boxed{4}\)[/tex] may have an issue in the provided options. However, given the exact and clear arithmetic simplification, the proper mathematical conclusion stands as [tex]\(4\)[/tex].
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.