`(0.bar1)^(2)[1-9(0.bar(16))^(2)]=?`

Options

A. `-1/(162)`
B. `1/(108)`
C. `7696/(106)`
D. `833/(88209)`

Answer

To solve the expression \((0.\overline{1})^2[1 - 9(0.\overline{16})^2]\), we will follow these steps:

### Step 1: Convert the repeating decimals to fractions
1. **Convert \(0.\overline{1}\)**:
Let \(x = 0.\overline{1}\).
Then, \(10x = 1.\overline{1}\).
Subtracting these gives:
\[ 10x - x = 1 \Rightarrow 9x = 1 \Rightarrow x = \frac{1}{9} \]
Thus, \(0.\overline{1} = \frac{1}{9}\).

2. **Convert \(0.\overline{16}\)**:
Let \(y = 0.\overline{16}\).
Then, \(100y = 16.\overline{16}\).
Subtracting gives:
\[ 100y - y = 16 \Rightarrow 99y = 16 \Rightarrow y = \frac{16}{99} \]
Thus, \(0.\overline{16} = \frac{16}{99}\).

### Step 2: Substitute the fractions into the expression
Now substitute these values into the original expression:
\[ (0.\overline{1})^2 = \left(\frac{1}{9}\right)^2 = \frac{1}{81} \]
\[ (0.\overline{16})^2 = \left(\frac{16}{99}\right)^2 = \frac{256}{9801} \]

### Step 3: Substitute into the equation
Now substitute these back into the expression:
\[ \frac{1}{81} \left[ 1 - 9 \left( \frac{256}{9801} \right) \right] \]

### Step 4: Simplify the expression inside the brackets
Calculate \(9 \times \frac{256}{9801}\):
\[ 9 \times \frac{256}{9801} = \frac{2304}{9801} \]
Now substitute this back:
\[ 1 - \frac{2304}{9801} = \frac{9801 - 2304}{9801} = \frac{7497}{9801} \]

### Step 5: Multiply by \(\frac{1}{81}\)
Now multiply:
\[ \frac{1}{81} \times \frac{7497}{9801} = \frac{7497}{81 \times 9801} \]
Calculating \(81 \times 9801\):
\[ 81 \times 9801 = 793581 \]
Thus, we have:
\[ \frac{7497}{793581} \]

### Final Result
The final result is:
\[ \frac{7497}{793581} \]