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Solve the following expressions:

1. [tex]\(\frac{1}{x-1}-\frac{2}{x^2-1}\)[/tex]

2. [tex]\(2^x+\frac{1}{2^x}=4 \frac{1}{4}\)[/tex]


Sagot :

Let's break down each expression separately using algebraic methods.

### Expression 1: [tex]\(\frac{1}{x-1} - \frac{2}{x^2-1}\)[/tex]

1. Factorize the denominator [tex]\(x^2-1\)[/tex]:
[tex]\[x^2 - 1 = (x-1)(x+1)\][/tex]

2. Rewrite the second term:
[tex]\[ \frac{2}{x^2-1} = \frac{2}{(x-1)(x+1)} \][/tex]

3. Find a common denominator for the terms:
The common denominator is [tex]\((x-1)(x+1)\)[/tex].

4. Rewrite the first term with the common denominator:
[tex]\[ \frac{1}{x-1} = \frac{x+1}{(x-1)(x+1)} \][/tex]

5. Combine the terms:
[tex]\[ \frac{x+1}{(x-1)(x+1)} - \frac{2}{(x-1)(x+1)} = \frac{x+1 - 2}{(x-1)(x+1)} = \frac{x-1}{(x-1)(x+1)} \][/tex]

6. Simplify the expression:
[tex]\[ \frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1} \quad \text{(since } x \neq 1\text{)} \][/tex]

So, the simplified form of the first expression is:
[tex]\[ \frac{1}{x+1} \][/tex]

### Expression 2: [tex]\(2^x + \frac{1}{2^x} = 4.25\)[/tex]

1. Let [tex]\(y = 2^x\)[/tex]:
[tex]\[ y + \frac{1}{y} = 4.25 \][/tex]

2. Multiply both sides by [tex]\(y\)[/tex] to eliminate the fraction:
[tex]\[ y^2 + 1 = 4.25y \][/tex]

3. Rearrange the equation to form a quadratic equation:
[tex]\[ y^2 - 4.25y + 1 = 0 \][/tex]

4. Solve the quadratic equation using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -4.25\)[/tex], and [tex]\(c = 1\)[/tex]:
[tex]\[ y = \frac{4.25 \pm \sqrt{4.25^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{4.25 \pm \sqrt{18.0625 - 4}}{2} \][/tex]
[tex]\[ y = \frac{4.25 \pm \sqrt{14.0625}}{2} \][/tex]
[tex]\[ y = \frac{4.25 \pm 3.75}{2} \][/tex]

5. Calculate the possible values of [tex]\(y\)[/tex]:
[tex]\[ y = \frac{4.25 + 3.75}{2} = 4 \quad \text{and} \quad y = \frac{4.25 - 3.75}{2} = 0.25 \][/tex]

6. Recall that [tex]\(y = 2^x\)[/tex], therefore:
[tex]\[ 2^x = 4 \quad \text{and} \quad 2^x = 0.25 \][/tex]

7. Solve for [tex]\(x\)[/tex]:
[tex]\[ 2^x = 4 \implies x = 2 \][/tex]
[tex]\[ 2^x = 0.25 \implies 2^x = 2^{-2} \implies x = -2 \][/tex]

So, the solutions for the second expression are:
[tex]\[ x = 2 \quad \text{and} \quad x = -2 \][/tex]

### Summary:
- The simplified form of [tex]\(\frac{1}{x-1} - \frac{2}{x^2-1}\)[/tex] is [tex]\(\frac{1}{x+1}\)[/tex].
- The solutions to [tex]\(2^x + \frac{1}{2^x} = 4.25\)[/tex] are [tex]\(x = 2\)[/tex] and [tex]\(x = -2\)[/tex].