To solve the given problem, we need to find the value of the expression \(\frac{(x + y)^2}{(x - y)^2}\) given that \(3x = 4y\).
Step 1: Express \(y\) in terms of \(x\)
Starting from the equation \(3x = 4y\):
[tex]\[\begin{aligned}
3x &= 4y \\
y &= \frac{3}{4}x
\end{aligned}\][/tex]
Step 2: Express \(x + y\) and \(x - y\) in terms of \(x\)
Now substitute \(y\) back into the expressions \(x + y\) and \(x - y\):
[tex]\[ x + y = x + \left(\frac{3}{4}x\right) = x + \frac{3}{4}x = \frac{4}{4}x + \frac{3}{4}x = \frac{7}{4}x \][/tex]
[tex]\[ x - y = x - \left(\frac{3}{4}x\right) = x - \frac{3}{4}x = \frac{4}{4}x - \frac{3}{4}x = \frac{1}{4}x \][/tex]
Step 3: Find the squares of these expressions
[tex]\[ (x + y)^2 = \left(\frac{7}{4}x\right)^2 = \left(\frac{7x}{4}\right)^2 = \frac{49x^2}{16} \][/tex]
[tex]\[ (x - y)^2 = \left(\frac{1}{4}x\right)^2 = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \][/tex]
Step 4: Determine the ratio of these squares
[tex]\[ \frac{(x + y)^2}{(x - y)^2} = \frac{\frac{49x^2}{16}}{\frac{x^2}{16}} = \frac{49x^2}{16} \times \frac{16}{x^2} \][/tex]
Since \( \frac{49x^2}{16} \times \frac{16}{x^2} = 49 \):
[tex]\[ \frac{(x + y)^2}{(x - y)^2} = 49 \][/tex]
Therefore, the value of [tex]\(\frac{(x + y)^2}{(x - y)^2}\)[/tex] is [tex]\(\boxed{49}\)[/tex].
