# Bewnwnwbbshsbsb

## Cumaali said

Let’s address the two math questions visible in the image step-by-step.

### Question 13:

[ x + \frac{1}{x} = 1 ]
[ \text{İşlem } \frac{25x^2 + 1}{x^2} \text{ ifadesinin değeri kaçtır?} ]

Answer:

We start by squaring the given equation:

[ (x + \frac{1}{x})^2 = 1^2 ]
[ x^2 + 2 \cdot \frac{1}{x} \cdot x + \frac{1}{x^2} = 1 ]
[ x^2 + 2 + \frac{1}{x^2} = 1 ]

Rearranging terms gives:

[ x^2 + \frac{1}{x^2} = 1 - 2 ]
[ x^2 + \frac{1}{x^2} = -1 ]

Next, we compute the expression ( \frac{25x^2 + 1}{x^2} ):

[ \frac{25x^2 + 1}{x^2} = 25 + \frac{1}{x^2} ]

Since we already found that ( x^2 + \frac{1}{x^2} = -1 ), this implies:

[ \frac{1}{x^2} = -1 - x^2 ]

But instead we directly get the value since ( x^2 + \frac{1}{x^2} = -1 ):

[ 25 + (-1) = 24 ]

So the value of the expression is ( \boxed{24} ).

### Question 14:

[ x + 2 \sqrt{x} = 6 ]
[ \text{İfadesinin değeri kaçtır \frac{x}{12} + \frac{1}{2} = ?} ]

Answer:

To solve for ( x ), we’ll start by isolating ( \sqrt{x} ):

[ \sqrt{x}(x + 2) = 6 ]

Square both sides:

[ x + 2 \sqrt{x} = 6 ]

Solving the quadratic-like equation:

Solve for ( x - 4 = 0 ):

^2 + 2 \sqrt{x} = 6
[ \sqrt{x
[x = 4 \LL - 2

The given expression in question is:

[ \frac{x}{12} + \frac{1}{2} ]

Substituting (x = 4 ) into this:

[ = \frac{x}{12}

And Compute each term:

[= \frac{ = 2) = \frac + 4 + \frac {1}{2 = 2}$So, the value of the expression is ( \boxed{12} + \frac$.

Hence the values would be (24.