This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions for the given expressions. Problem 1: (1)/(2(x-2)) + (1)/(6(x-2)) - (1)/(3(x-1)) Step 1: Find the least common denominator (LCD). The denominators are 2(x-2), 6(x-2), and 3(x-1). The LCD is 6(x-2)(x-1). Step 2: Rewrite each fraction with the LCD. (1 · 3(x-1))/(2(x-2) · 3(x-1)) + (1 · (x-1))/(6(x-2) · (x-1)) - (1 · 2(x-2))/(3(x-1) · 2(x-2)) = (3(x-1))/(6(x-2)(x-1)) + (x-1)/(6(x-2)(x-1)) - (2(x-2))/(6(x-2)(x-1)) Step 3: Combine the numerators over the common denominator. = (3(x-1) + (x-1) - 2(x-2))/(6(x-2)(x-1)) Step 4: Simplify the numerator. = (3x - 3 + x - 1 - 2x + 4)/(6(x-2)(x-1)) = ((3x + x - 2x) + (-3 - 1 + 4))/(6(x-2)(x-1)) = (2x + 0)/(6(x-2)(x-1)) = (2x)/(6(x-2)(x-1)) Step 5: Reduce the fraction. = (x)/(3(x-2)(x-1)) The simplified expression is: (x)/(3(x-2)(x-1)) Problem 2: (1)/(x+1) + (2)/((x-1)^2) - (x^2)/((x-1)^3) Step 1: Find the least common denominator (LCD). The denominators are (x+1), (x-1)^2, and (x-1)^3. The LCD is (x+1)(x-1)^3. Step 2: Rewrite each fraction with the LCD. (1 · (x-1)^3)/((x+1)(x-1)^3) + (2 · (x+1)(x-1))/((x-1)^2 · (x+1)(x-1)) - (x^2 · (x+1))/((x-1)^3 · (x+1)) = ((x-1)^3)/((x+1)(x-1)^3) + (2(x^2-1))/((x+1)(x-1)^3) - (x^2(x+1))/((x+1)(x-1)^3) Step 3: Combine the numerators over the common denominator. = ((x-1)^3 + 2(x^2-1) - x^2(x+1))/((x+1)(x-1)^3) Step 4: Expand and simplify the numerator. Recall (x-1)^3 = x^3 - 3x^2 + 3x - 1. = ((x^3 - 3x^2 + 3x - 1) + (2x^2 - 2) - (x^3 + x^2))/((x+1)(x-1)^3) = (x^3 - 3x^2 + 3x - 1 + 2x^2 - 2 - x^3 - x^2)/((x+1)(x-1)^3) Combine like terms: = ((x^3 - x^3) + (-3x^2 + 2x^2 - x^2) + (3x) + (-1 - 2))/((x+1)(x-1)^3) = (0 - 2x^2 + 3x - 3)/((x+1)(x-1)^3) = (-2x^2 + 3x - 3)/((x+1)(x-1)^3) The simplified expression is: (-2x^2 + 3x - 3)/((x+1)(x-1)^3) Send me the next one 📸