26.If $$ \frac{1+\sqrt{2}}{1-\sqrt{2}}+\frac{1-\sqrt{2}}{1+\sqrt{2}}=a+b \sqrt{2} $$, Find the value of $$ a $$ and $$ b $$

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UpStudy AI · Accepted Answer

To solve the expression:

$$
\frac{1+\sqrt{2}}{1-\sqrt{2}} + \frac{1-\sqrt{2}}{1+\sqrt{2}} = a + b\sqrt{2}
$$

we can simplify each fraction separately and then combine the results.

1. **Rationalize Each Fraction:**

- For the first fraction $$\frac{1+\sqrt{2}}{1-\sqrt{2}}$$, multiply the numerator and denominator by the conjugate of the denominator:

$$
     \frac{1+\sqrt{2}}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} = \frac{(1+\sqrt{2})^2}{1 - (\sqrt{2})^2} = \frac{1 + 2\sqrt{2} + 2}{1 - 2} = \frac{3 + 2\sqrt{2}}{-1} = -3 - 2\sqrt{2}
     $$

- For the second fraction $$\frac{1-\sqrt{2}}{1+\sqrt{2}}$$, similarly multiply by the conjugate:

$$
     \frac{1-\sqrt{2}}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{(1-\sqrt{2})^2}{1 - (\sqrt{2})^2} = \frac{1 - 2\sqrt{2} + 2}{1 - 2} = \frac{3 - 2\sqrt{2}}{-1} = -3 + 2\sqrt{2}
     $$

2. **Combine the Results:**

$$
   -3 - 2\sqrt{2} + (-3 + 2\sqrt{2}) = -6
   $$

The $$\sqrt{2}$$ terms cancel out, leaving only $$-6$$.

3. **Conclusion:**

Comparing the simplified expression to $$a + b\sqrt{2}$$, we see that:

$$
   a = -6 \quad \text{and} \quad b = 0
   $$

**Final Answer:**
$$a = -6$$ and $$b = 0$$
$$a = -6$$ and $$b = 0$$.

26.If , Find the value of and

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