Square of Difference: Using fractions

To solve this problem, we'll use the square of a binomial formula:

• Identify $a = \frac{1}{x}$ and $b = x$.
• Apply the formula $(a - b)^2 = a^2 - 2ab + b^2$.
• Simplify each component and obtain the final expression.

Let's work through the steps:

Step 1: Identify the components:
$a = \frac{1}{x}$, $b = x$

Step 2: Apply the formula:

$(\frac{1}{x} - x)^2 = (\frac{1}{x})^2 - 2 \times \frac{1}{x} \times x + x^2$

Simplifying each term:

$(\frac{1}{x})^2 = \frac{1}{x^2}$

$-2 \times \frac{1}{x} \times x = -2$

$x^2 = x^2$

Step 3: Combine and simplify:

$\frac{1}{x^2} - 2 + x^2$

To combine these into a single fraction, find a common denominator $x^2$:

$\frac{1}{x^2} - \frac{2x^2}{x^2} + \frac{x^4}{x^2} = \frac{1 - 2x^2 + x^4}{x^2}$

Thus, the simplified expression is:

$\frac{x^4 - 2x^2 + 1}{x^2}$

Comparing to the choices provided, the correct choice is:

Choice 3: $\frac{x^4 - 2x^2 + 1}{x^2}$

To solve this problem, we'll begin by simplifying the given expression:

The equation is $\frac{x}{4y} + \frac{y}{x} - 1 = 0$.

• Step 1: Eliminate the fraction by obtaining a common denominator for the left side of the equation. The common denominator is $4y \times x = 4xy$.
• Step 2: Rewrite each term with the common denominator:
  $\frac{x^2}{4xy} + \frac{4y^2}{4xy} - \frac{4y}{4y} = 0$. This becomes $\frac{x^2 + 4y^2 - 4xy}{4xy} = 0$.
• Step 3: Multiply through by $4xy$ to eliminate the denominator: $x^2 + 4y^2 - 4xy = 0$.
• Step 4: Recognize this as being equivalent to the expanded form of $(x - 2y)^2 = x^2 - 2xy + 4y^2 = 0$.
• Step 5: Confirm that $x^2 - 2xy + 4y^2 \equiv x^2 + 4y^2 - 4xy$, indicating $(x - 2y)^2 = 0$.

Thus, we conclude that the equation can be rewritten using the square of a difference formula:

Therefore, the short multiplication formula for the given equation is $(x - 2y)^2 = 0$.

To solve this problem, we'll follow these steps:

• Step 1: Simplify the expression inside the square root.
• Step 2: Utilize the algebraic identity to combine and simplify terms.
• Step 3: Divide the result by $x-y$.

Now, let's work through each step:

Step 1: Expand $(x-y)^2$.

We have $(x-y)^2 = x^2 - 2xy + y^2$.

Step 2: Simplify the square root expression:
The expression inside the square root is:
$2x^2 - 4xy + 2y^2 + (x-y)^2$.
Substitute $(x-y)^2 = x^2 - 2xy + y^2$:
$2x^2 - 4xy + 2y^2 + x^2 - 2xy + y^2$.

This simplifies to:
$3x^2 - 6xy + 3y^2$.

Notice that this can be rewritten using the identity $(a-b)^2 = a^2 - 2ab + b^2$ as:
$3(x^2 - 2xy + y^2) = 3(x-y)^2$.

Step 3: Extract the square root and simplify:
$\sqrt{3(x-y)^2} = \sqrt{3} \times |x-y|$.

Finally, divide by $x-y$:
$\frac{\sqrt{3} \times |x-y|}{x-y} = \sqrt{3} \times \frac{|x-y|}{x-y}$.

Since we assume $x-y \neq 0$, it simplifies to $\sqrt{3}$ because $\frac{|x-y|}{x-y} = 1$ when $x > y$, and $-1$ when $x < y$. With the absolute value, it remains $1$ in both cases.

Therefore, the solution to the problem is $\sqrt{3}$.

To solve this problem, we'll follow these steps:

• Step 1: Substitute the given value of $(x-y)^2$ into the first equation.
• Step 2: Use the identity for a square of a difference to find a relationship between $x^2 + y^2$ and $xy$.
• Step 3: Solve for the product $xy$.

Now, let's work through each step:

Step 1: We start with the provided equation:

$\frac{x^2 + y^2}{(x-y)^2} = 3$

Given that $(x-y)^2 = 1$, we substitute:

$\frac{x^2 + y^2}{1} = 3$

which simplifies to:

$x^2 + y^2 = 3$

Step 2: We know from the identity of a square of a difference:

$(x-y)^2 = x^2 - 2xy + y^2$

Given $(x-y)^2 = 1$, we can write:

$x^2 - 2xy + y^2 = 1$

Step 3: We set up a system of equations:

$x^2 + y^2 = 3$ (Equation 1)

$x^2 - 2xy + y^2 = 1$ (Equation 2)

Subtract Equation 2 from Equation 1:

$(x^2 + y^2) - (x^2 - 2xy + y^2) = 3 - 1$

Simplifying the left side gives $2xy$:

$2xy = 2$

Divide both sides by 2:

$xy = 1$

Therefore, the product of $x$ and $y$ is $xy = 1$.

To solve this problem, we'll follow these steps:

• Step 1: Expand and simplify the numerator $(\frac{1}{x} - \frac{1}{2})^2$.
• Step 2: Expand and simplify the denominator $(\frac{1}{x} - \frac{1}{3})^2$.
• Step 3: Set up the equation as a proportion and solve for $x$.

Let’s work through each step:
Step 1: Using the formula for the square of a difference, expand the numerator:

$(\frac{1}{x} - \frac{1}{2})^2 = \left(\frac{1}{x}\right)^2 - 2\left(\frac{1}{x}\right)\left(\frac{1}{2}\right) + \left(\frac{1}{2}\right)^2 = \frac{1}{x^2} - \frac{1}{x} + \frac{1}{4}$.

Step 2: Similarly, expand the denominator:

$(\frac{1}{x} - \frac{1}{3})^2 = \left(\frac{1}{x}\right)^2 - 2\left(\frac{1}{x}\right)\left(\frac{1}{3}\right) + \left(\frac{1}{3}\right)^2 = \frac{1}{x^2} - \frac{2}{3x} + \frac{1}{9}$.

Step 3: Substitute these into the original equation and solve the proportion:

$\frac{\frac{1}{x^2} - \frac{1}{x} + \frac{1}{4}}{\frac{1}{x^2} - \frac{2}{3x} + \frac{1}{9}} = \frac{9}{4}$.

Cross-multiply to clear the fractions:

$4\left(\frac{1}{x^2} - \frac{1}{x} + \frac{1}{4}\right) = 9\left(\frac{1}{x^2} - \frac{2}{3x} + \frac{1}{9}\right)$.

Simplifying both sides gives:

$4(\frac{1}{x^2} - \frac{1}{x} + \frac{1}{4}) = 4\frac{1}{x^2} - 4\frac{1}{x} + 1$.

$9(\frac{1}{x^2} - \frac{2}{3x} + \frac{1}{9}) = 9\frac{1}{x^2} - 6\frac{1}{x} + 1$.

Equating the expressions, we have:

$4\frac{1}{x^2} - 4\frac{1}{x} + 1 = 9\frac{1}{x^2} - 6\frac{1}{x} + 1$.

Subtract 1 from both sides and collect like terms:

$-4\frac{1}{x} + 1 = 5\frac{1}{x^2} - 2\frac{1}{x}$.

$-2\frac{1}{x} - 5\frac{1}{x^2} = 0$.

Factoring gives:

$5\frac{1}{x}(x - 2) = 0$.

Therefore, the solution for $x$ should satisfy $x - 2 = 0$, so $x = 2.5$.

Thus, the value of $x$ is $\boxed{2.5}$.

To solve this problem, we'll proceed with the following steps:

• Step 1: Expand the expression $(2x-3)^2$ on the right-hand side.
• Step 2: Simplify the resulting expression by dividing by $x$.
• Step 3: Compare the terms on both sides of the equation to deduce the coefficients $A$, $B$, and $C$.

Now, let us work through these steps:

Step 1: Expand the expression $(2x-3)^2$.

$(2x - 3)^2 = 4x^2 - 12x + 9$

Step 2: Substitute back into the equation and simplify by dividing by $x$:

$\frac{A}{x} + \frac{Bx}{2} = \frac{4x^2 - 12x + 9}{x} - C$

This simplifies to:

$4x - 12 + \frac{9}{x} - C$

Step 3: To find $A$, $B$, and $C$, compare coefficients of similar terms:

• Constant term (no $x$): Equate $-12$ with $-C$ to get $C = 12$ (note: should consider sign reversal).
• Coefficient of $x$: Compare $\frac{B}{2}$ to 4. Giving $B = 8$.
• Coefficient of $\frac{1}{x}$: Compare $A$ to 9, giving $A = 9$.

Therefore, the solution to the problem is that the values are $\textbf{A = 9, B = 8, C = -12}$.