If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to :

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For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.

Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:

$5 + 5r > 5r^2$
$5 + 5r^2 > 5r$
$5r + 5r^2 > 5$

Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.

1. $1 + r > r^2$

We can rewrite this as:
$r^2 - r - 1 < 0$

$\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]

2. $1 + r\^2 > r$

We can rewrite this as:
$r\^2 - r + 1 > 0$

$$
\begin{gathered}  \Rightarrow  r^2-2 \cdot \frac{1}{2} r+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2>0 \ \Rightarrow  \left(r-\frac{1}{2}\right)^2+\frac{3}{4}>0 \ \Rightarrow  r \in R \end{gathered}
$$

This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.

3. $r + r\^2 > 1$

We can rewrite this as:
$r\^2 + r - 1 > 0$

$$
\begin{aligned}

& \Rightarrow {\left[r-\left(\frac{-1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{-1+\sqrt{5}}{2}\right)\right]>0} \\
& {\left[\because r^2+r-1=0 \Rightarrow r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}\right]} \\
& \Rightarrow r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)
\end{aligned}
$$

Taking the intersection of the intervals from the three inequalities, we get $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$.

Given the solution of the inequalities $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$, we need to determine which of the options are outside this interval.

Option A: $\frac{7}{4}$

The decimal representation of $\frac{7}{4}$ is 1.75, which is outside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\frac{7}{4}$.

Option B: $\frac{5}{4}$

The decimal representation of $\frac{5}{4}$ is 1.25, which is inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{5}{4}$.

Option C: $\frac{3}{4}$

The decimal representation of $\frac{3}{4}$ is 0.75, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{4}$.

Option D: $\frac{3}{2}$

The decimal representation of $\frac{3}{2}$ is 1.5, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{2}$.

Therefore, the only value that r cannot be equal to, among the given options, is $\frac{7}{4}$. So, the correct answer is:

Option A: $

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