a \( \sqrt{75}+\sqrt{27}-\sqrt{12} \) b \( \frac{11^{n+1} 10^{n-1}}{2^{n-1} 55^{n+1}} \) Solve for \( x \)

It appears you have two separate problems. (There is no variable x in the expressions you provided.) I’ll simplify each one step‐by‐step. ────────────────────────────── Problem (a): Simplify √75 + √27 − √12. Step 1. Write each square root in simplified form by factoring out perfect squares: • √75 = √(25·3) = √25 · √3 = 5√3 • √27 = √(9·3) = √9 · √3 = 3√3 • √12 = √(4·3) = √4 · √3 = 2√3 Step 2. Substitute these into the expression:   √75 + √27 − √12   = 5√3 + 3√3 − 2√3 Step 3. Combine like terms (all terms are multiples of √3):   5√3 + 3√3 − 2√3 = (5 + 3 − 2)√3 = 6√3 Thus, the simplified result for (a) is 6√3. ────────────────────────────── Problem (b): Simplify (11^(n+1) · 10^(n−1)) / (2^(n−1) · 55^(n+1)). Step 1. Express 10 and 55 in terms of their prime factors:   10 = 2 · 5   55 = 5 · 11 Step 2. Rewrite the expression using these factors:   10^(n−1) = (2·5)^(n−1) = 2^(n−1) · 5^(n−1)   55^(n+1) = (5·11)^(n+1) = 5^(n+1) · 11^(n+1) Therefore, the expression becomes:   [11^(n+1) · (2^(n−1) · 5^(n−1))] / [2^(n−1) · (5^(n+1) · 11^(n+1))]. Step 3. Cancel common factors:  • The factor 11^(n+1) in the numerator and denominator cancels.  • The factor 2^(n−1) cancels as well. This leaves:   5^(n−1) / 5^(n+1). Step 4. Simplify using the law of exponents (a^(m)/a^(n) = a^(m−n)):   5^(n−1) / 5^(n+1) = 5^[(n−1) − (n+1)] = 5^(−2) = 1/5^2 = 1/25. Thus, the simplified result for (b) is 1/25. ────────────────────────────── Conclusion: a) √75 + √27 − √12 = 6√3 b) (11^(n+1) · 10^(n−1)) / (2^(n−1) · 55^(n+1)) = 1/25 If you intended "Solve for x" in another context or if there is an equation involving x that was not provided, please clarify so I can help further.