Square of sum: Using fractions

To solve this problem, we'll use the formula for the square of a sum.

Let's define our terms:
Let $a = \frac{1}{x}$ and $b = x$.

According to the formula $(a + b)^2 = a^2 + 2ab + b^2$, we need to find the following:
1. $a^2 = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}$
2. $2ab = 2 \times \frac{1}{x} \times x = 2$
3. $b^2 = x^2$

Substituting these into the formula gives us:

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

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

• Convert $2$ to a fraction with $x^2$ as the denominator: $2 = \frac{2x^2}{x^2}$
• Convert $x^2$ to a fraction with $x^2$ as the denominator: $x^2 = \frac{x^4}{x^2}$

So, the expression becomes:

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

Therefore, the expanded expression is $\frac{x^4 + 2x^2 + 1}{x^2}$.

To solve this problem, let's work through these steps:

• Step 1: Start with the given equation: $1+\frac{y}{x}+\frac{x}{4y}=0$.
• Step 2: Clear the fractions by multiplying all terms by $4xy$, the common denominator.
• Step 3: This gives us: $4xy + 4y^2 + x^2 = 0$.
• Step 4: Rearrange the terms for clarity: $x^2 + 4xy + 4y^2 = 0$.
• Step 5: Recognize this as the perfect square: $(x + 2y)^2 = 0$.

This simplification results in the equation: $(x+2y)^2 = 0$.

Therefore, the solution to the problem is $(x+2y)^2 = 0$.

To solve this problem, we will simplify both sides of the given equation:

Given equation:

$\frac{A}{X} + \frac{BX}{2} = \frac{(2X+3)^2}{X} - C$.

First, expand the quadratic expression:

$(2X+3)^2 = (2X+3)(2X+3) = 4X^2 + 6X + 6X + 9 = 4X^2 + 12X + 9$.

Substitute this back into the equation:

$\frac{A}{X} + \frac{BX}{2} = \frac{4X^2 + 12X + 9}{X} - C$.

Simplify the right-hand side:

$\frac{4X^2 + 12X + 9}{X} = 4X + \frac{12X}{X} + \frac{9}{X} = 4X + 12 + \frac{9}{X}$.

The equation now becomes:

$\frac{A}{X} + \frac{BX}{2} = 4X + 12 + \frac{9}{X} - C$.

For the equation to hold true for all values of $X$, equate corresponding terms:

• $\frac{A}{X} = \frac{9}{X} \Rightarrow A = 9$.
• $\frac{BX}{2} = 4X \Rightarrow B = 8$.
• For constant terms: $12 - C = 0 \Rightarrow C = 12$.

Therefore, the values are $A = 9$, $B = 8$, and $C = 12$.

The correct answer is: $A = 9, B = 8, C = 12$.

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: Cross-multiply the given equation to eliminate fractions.
• Step 2: Expand the squared terms on either side of the equation.
• Step 3: Rearrange terms to form a quadratic equation.
• Step 4: Solve the quadratic equation to find possible values for $x$.

Now, let's work through each step:

Step 1: Begin with the given equation:
$\frac{(\frac{1}{x} + \frac{1}{2})^2}{(\frac{1}{x} + \frac{1}{3})^2} = \frac{81}{64}$. Cross-multiply to eliminate fractions:
$(\frac{1}{x} + \frac{1}{2})^2 \times 64 = (\frac{1}{x} + \frac{1}{3})^2 \times 81$.

Step 2: Expand each squared term:
For $(\frac{1}{x} + \frac{1}{2})^2$, use $(a + b)^2 = a^2 + 2ab + b^2$:
$(\frac{1}{x})^2 + 2(\frac{1}{x})(\frac{1}{2}) + (\frac{1}{2})^2 = \frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}$.
Similarly, $(\frac{1}{x} + \frac{1}{3})^2 = \frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}$.

Step 3: Substitute these into the cross-multiplied equation:
$64\left(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}\right) = 81\left(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}\right)$.

Step 4: Simplify and collect like terms:
$64(\frac{1}{x^2} + \frac{1}{x} + \frac{1}{4}) = (64\frac{1}{x^2} + 64\frac{1}{x} + 16)$,
$81(\frac{1}{x^2} + \frac{2}{3x} + \frac{1}{9}) = (81\frac{1}{x^2} + 54\frac{1}{x} + 9)$.

Equating terms gives:
$64\frac{1}{x^2} + 64\frac{1}{x} + 16 = 81\frac{1}{x^2} + 54\frac{1}{x} + 9$.

Step 5: Solve the quadratic equation:
Combine like terms: $-17\frac{1}{x^2} + 10\frac{1}{x} + 7 = 0$.
Let $y = \frac{1}{x}$. Substitute to get: $-17y^2 + 10y + 7 = 0$.
Multiply the entire equation by -1 to simplify: $17y^2 - 10y - 7 = 0$.

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=17$, $b=-10$, $c=-7$:
$y = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 17 \times (-7)}}{2 \times 17}$
$y = \frac{10 \pm \sqrt{100 + 476}}{34}$
$y = \frac{10 \pm \sqrt{576}}{34}$
$y = \frac{10 \pm 24}{34}$
Which gives:
$y = \frac{34}{34} = 1$ or $y = -\frac{14}{34} = -\frac{7}{17}$.

Since $y = \frac{1}{x}$:
For $y=1$, $x = \frac{1}{y} = 1$.
For $y=-\frac{7}{17}$, $x = \frac{1}{y} = -\frac{17}{7}$.

Therefore, the solutions for $x$ are $x = 1$ and $x = -\frac{17}{7}$.

Checking the correct answer choice, these correspond to the second choice.

Thus, the solution to the problem is $x = 1, -\frac{17}{7}$.

To solve the given problem, we will proceed with the following steps:

• Step 1: Apply the identity for the square of the sum.
• Step 2: Use the given equation to solve for the variables.
• Step 3: Derive the product $xy$.

Step 1: Using the identity
For $(x+y)^2 = x^2 + 2xy + y^2$, we know from the problem that $(x+y)^2 = 1$, so:

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

Step 2: Utilizing the second given equation,
We have $\frac{x^2+y^2}{(x+y)^2}=3$. Therefore:

$x^2 + y^2 = 3(x+y)^2 = 3 \times 1 = 3$

Step 3: Substitute $x^2+y^2 = 3$ into the identity:

$3 + 2xy = 1$

Solving for $xy$, we get:

$2xy = 1 - 3 = -2$

$xy = -1$

Therefore, the product of $x$ and $y$ is $\mathbf{-1}$.