Simplify. \( -\frac{5-3 e}{4-2 e} \pm \sqrt{\left(\frac{5-3 e}{4-2 e}\right)^{2}-\frac{3}{e-2}} \)

We start with the expression \[ -\frac{5-3e}{4-2e} \pm \sqrt{\left(\frac{5-3e}{4-2e}\right)^2-\frac{3}{e-2}}. \] **Step 1. Rewrite the fraction** Notice that \[ 4-2e=2(2-e). \] Since \[ 2-e=-(e-2), \] we have \[ \frac{5-3e}{4-2e}=\frac{5-3e}{2(2-e)}=\frac{5-3e}{2(-1)(e-2)}=-\frac{5-3e}{2(e-2)}. \] Then the whole expression becomes \[ -\Bigl(-\frac{5-3e}{2(e-2)}\Bigr) \pm \sqrt{\left(-\frac{5-3e}{2(e-2)}\right)^2-\frac{3}{e-2}} =\frac{5-3e}{2(e-2)} \pm \sqrt{\frac{(5-3e)^2}{4(e-2)^2}-\frac{3}{e-2}}. \] **Step 2. Simplify the expression inside the square root** Write the radicand as a single fraction: \[ \frac{(5-3e)^2}{4(e-2)^2}-\frac{3}{e-2} =\frac{(5-3e)^2 - 12(e-2)}{4(e-2)^2}. \] Expand and simplify the numerator: \[ (5-3e)^2 = 25 - 30e + 9e^2. \] Then \[ 25-30e+9e^2 - 12(e-2)=25-30e+9e^2 - 12e + 24 = 9e^2-42e+49. \] Notice that \[ 9e^2-42e+49=(3e-7)^2. \] Thus, the radicand becomes \[ \frac{(3e-7)^2}{4(e-2)^2}. \] Taking the square root gives \[ \sqrt{\frac{(3e-7)^2}{4(e-2)^2}}=\frac{|3e-7|}{2|e-2|}. \] For simplicity (and assuming an appropriate sign choice in context), we write the result as \[ \frac{3e-7}{2(e-2)} \] so that the whole expression becomes \[ \frac{5-3e}{2(e-2)} \pm \frac{3e-7}{2(e-2)}. \] **Step 3. Combine the fractions** Since the two terms now have a common denominator, we combine them: \[ \frac{5-3e \pm (3e-7)}{2(e-2)}. \] This gives two cases: 1. With the \(+\) sign: \[ \frac{5-3e +3e-7}{2(e-2)}=\frac{5-7}{2(e-2)}=\frac{-2}{2(e-2)}=-\frac{1}{e-2}. \] 2. With the \(-\) sign: \[ \frac{5-3e - (3e-7)}{2(e-2)}=\frac{5-3e-3e+7}{2(e-2)} =\frac{12-6e}{2(e-2)} =\frac{6(2-e)}{2(e-2)}. \] Notice that \[ 2-e=-(e-2), \] so \[ \frac{6(2-e)}{2(e-2)}=\frac{6(-1)(e-2)}{2(e-2)}=-\frac{6}{2}=-3. \] **Final Answer** Thus, the given expression simplifies (under our assumed sign conventions) to \[ \frac{5-3e \pm (3e-7)}{2(e-2)}=\begin{cases} -\frac{1}{e-2} & \text{(with the plus sign)},\\[1mm] -3 & \text{(with the minus sign)}. \end{cases} \]