https://r2.erweima.ai/mathbot/image/1719230080255acwxfat2.jpg

Let's re-evaluate the provided expression with the correct values from the latest image. The expression is:

\[
\frac{15(21086) - (514)(429)}{\sqrt{[15(26550) - (514)^2] \times [15(26191) - (429)^2]}}
\]

### Step 1: Evaluate the Numerator

\[
15 \times 21086 - 514 \times 429
\]

First, calculate each part:

\[
15 \times 21086 = 316290
\]
\[
514 \times 429 = 220506
\]

Then subtract the results:

\[
316290 - 220506 = 95784
\]

### Step 2: Evaluate the Denominator

The denominator consists of two parts inside the square root which are then multiplied:

\[
\sqrt{(15 \times 26550 - 514^2) \times (15 \times 26191 - 429^2)}
\]

First, calculate each part inside the parentheses:

\[
15 \times 26550 = 398250
\]
\[
514^2 = 264196
\]
\[
15 \times 26191 = 392865
\]
\[
429^2 = 184041
\]

Now subtract these results:

\[
398250 - 264196 = 134054
\]
\[
392865 - 184041 = 208824
\]

Then multiply these two results:

\[
134054 \times 208824 = 27998811496
\]

Finally, take the square root of the product:

\[
\sqrt{27998811496} \approx 167329.47
\]

### Step 3: Calculate the Final Result

Divide the evaluated numerator by the evaluated denominator:

\[
\frac{95784}{167329.47} \approx 0.5725
\]

Thus, the value of the given expression is approximately \(0.5725\).

Would you like more details or have any questions about this calculation? Here are five more questions you might want to ask:

1. How do you simplify complex fractions?
2. Can you explain the steps for solving quadratic equations?
3. What are the properties of logarithms?
4. How do you find the derivatives of trigonometric functions?
5. What is the process for solving systems of linear equations?

**Tip:** Always ensure to use precise values when performing mathematical operations to avoid errors in the final result.

Learn how to evaluate a complex fraction involving algebraic operations and square roots step-by-step. This problem covers algebra, fraction operations, and square root operations, providing detailed calculations and solutions. Suitable for students in Grades 10-12.

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