Add: `(3x^(2) - (1)/(5)x + (7)/(3)) + ((-1)/(4)x^(2) + (1)/(3)x - (1)/(6)) + (-2x^(2) - (1)/(2)x + 5)`

Answer

To solve the problem of adding the three polynomials, we will follow these steps:

### Step 1: Write down the polynomials
We have the following three polynomials to add:
1. \( 3x^2 - \frac{1}{5}x + \frac{7}{3} \)
2. \( -\frac{1}{4}x^2 + \frac{1}{3}x - \frac{1}{6} \)
3. \( -2x^2 - \frac{1}{2}x + 5 \)

### Step 2: Group the like terms
We will group the terms based on their degrees (i.e., \(x^2\), \(x\), and constant terms):
- **Quadratic terms**: \(3x^2 - \frac{1}{4}x^2 - 2x^2\)
- **Linear terms**: \(-\frac{1}{5}x + \frac{1}{3}x - \frac{1}{2}x\)
- **Constant terms**: \(\frac{7}{3} - \frac{1}{6} + 5\)

### Step 3: Add the quadratic terms
To add the quadratic terms:
\[ 3 - \frac{1}{4} - 2 = 3 - 2 - \frac{1}{4} = 1 - \frac{1}{4} = \frac{4}{4} - \frac{1}{4} = \frac{3}{4} \]
Thus, the combined quadratic term is:
\[ \frac{3}{4}x^2 \]

### Step 4: Add the linear terms
To add the linear terms, we need a common denominator. The least common multiple (LCM) of 5, 3, and 2 is 30:
\[ -\frac{1}{5} = -\frac{6}{30}, \quad \frac{1}{3} = \frac{10}{30}, \quad -\frac{1}{2} = -\frac{15}{30} \]
Now, adding these:
\[ -\frac{6}{30} + \frac{10}{30} - \frac{15}{30} = \frac{-6 + 10 - 15}{30} = \frac{-11}{30} \]

### Step 5: Add the constant terms
For the constant terms, we also find a common denominator. The LCM of 3, 6, and 1 is 6:
\[ \frac{7}{3} = \frac{14}{6}, \quad -\frac{1}{6} = -\frac{1}{6}, \quad 5 = \frac{30}{6} \]
Now, adding these:
\[ \frac{14}{6} - \frac{1}{6} + \frac{30}{6} = \frac{14 - 1 + 30}{6} = \frac{43}{6} \]

### Step 6: Combine all the results
Now we can combine all the results:
\[ \frac{3}{4}x^2 - \frac{11}{30}x + \frac{43}{6} \]

### Final Answer
Thus, the final result of the addition of the three polynomials is:
\[ \frac{3}{4}x^2 - \frac{11}{30}x + \frac{43}{6} \]

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