(10) \( \frac{4}{(2 x+1)^{2}}-\frac{x+1}{2 x+1} \) (12) \( \frac{x-3}{3}-\frac{x-2}{2}+\frac{x+1}{6} \)

Simplify the expression by following steps: - step0: Solution: \(\frac{\left(x-3\right)}{3}-\frac{\left(x-2\right)}{2}+\frac{\left(x+1\right)}{6}\) - step1: Remove the parentheses: \(\frac{x-3}{3}-\frac{x-2}{2}+\frac{x+1}{6}\) - step2: Reduce fractions to a common denominator: \(\frac{\left(x-3\right)\times 2}{3\times 2}-\frac{\left(x-2\right)\times 3}{2\times 3}+\frac{x+1}{6}\) - step3: Multiply the numbers: \(\frac{\left(x-3\right)\times 2}{6}-\frac{\left(x-2\right)\times 3}{2\times 3}+\frac{x+1}{6}\) - step4: Multiply the numbers: \(\frac{\left(x-3\right)\times 2}{6}-\frac{\left(x-2\right)\times 3}{6}+\frac{x+1}{6}\) - step5: Transform the expression: \(\frac{\left(x-3\right)\times 2-\left(x-2\right)\times 3+x+1}{6}\) - step6: Multiply the terms: \(\frac{2x-6-\left(x-2\right)\times 3+x+1}{6}\) - step7: Multiply the terms: \(\frac{2x-6-\left(3x-6\right)+x+1}{6}\) - step8: Calculate: \(\frac{1}{6}\) Calculate or simplify the expression \( 4/(2*x+1)^2 - (x+1)/(2*x+1) \). Simplify the expression by following steps: - step0: Solution: \(\frac{4}{\left(2x+1\right)^{2}}-\frac{\left(x+1\right)}{\left(2x+1\right)}\) - step1: Remove the parentheses: \(\frac{4}{\left(2x+1\right)^{2}}-\frac{x+1}{2x+1}\) - step2: Reduce fractions to a common denominator: \(\frac{4}{\left(2x+1\right)^{2}}-\frac{\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x+1\right)}\) - step3: Multiply: \(\frac{4}{\left(2x+1\right)^{2}}-\frac{\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^{2}}\) - step4: Transform the expression: \(\frac{4-\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)^{2}}\) - step5: Multiply the terms: \(\frac{4-\left(2x^{2}+3x+1\right)}{\left(2x+1\right)^{2}}\) - step6: Subtract the terms: \(\frac{3-2x^{2}-3x}{\left(2x+1\right)^{2}}\) - step7: Calculate: \(\frac{3-2x^{2}-3x}{4x^{2}+4x+1}\) Let's simplify both expressions step by step. ### Expression (10) The expression is: \[ \frac{4}{(2x+1)^{2}} - \frac{x+1}{2x+1} \] 1. **Finding a common denominator**: The common denominator for the two fractions is \((2x+1)^{2}\). 2. **Rewriting the second fraction**: \[ \frac{x+1}{2x+1} = \frac{(x+1)(2x+1)}{(2x+1)^{2}} = \frac{(x+1)(2x+1)}{(2x+1)^{2}} \] 3. **Combining the fractions**: \[ \frac{4 - (x+1)(2x+1)}{(2x+1)^{2}} \] 4. **Expanding the numerator**: \[ (x+1)(2x+1) = 2x^2 + 3x + 1 \] So, the numerator becomes: \[ 4 - (2x^2 + 3x + 1) = 3 - 2x^2 - 3x \] 5. **Final result**: \[ \frac{3 - 2x^{2} - 3x}{(2x+1)^{2}} \] ### Expression (12) The expression is: \[ \frac{x-3}{3} - \frac{x-2}{2} + \frac{x+1}{6} \] 1. **Finding a common denominator**: The common denominator for the three fractions is \(6\). 2. **Rewriting each fraction**: \[ \frac{x-3}{3} = \frac{2(x-3)}{6} = \frac{2x - 6}{6} \] \[ \frac{x-2}{2} = \frac{3(x-2)}{6} = \frac{3x - 6}{6} \] \[ \frac{x+1}{6} = \frac{x+1}{6} \] 3. **Combining the fractions**: \[ \frac{(2x - 6) - (3x - 6) + (x + 1)}{6} \] 4. **Simplifying the numerator**: \[ 2x - 6 - 3x + 6 + x + 1 = 0x + 1 = 1 \] 5. **Final result**: \[ \frac{1}{6} \] ### Summary of Results - For expression (10): \[ \frac{3 - 2x^{2} - 3x}{(2x+1)^{2}} \] - For expression (12): \[ \frac{1}{6} \]