Domain: Multiple domains

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$9(4x-\frac{5}{x})=20(3x-\frac{6}{x+1})$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation,

These fractions are considered defined as long as the expressions in their denominators are not equal to zero (since division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as shown below:

For the fraction inside of the parentheses in the expression on the left side we obtain the following:

$\boxed{ x\neq0}$

For the fraction inside of the parentheses in the expression on the right side we obtain the following:

$x+1\neq0 \\$Proceed to solve the second inequality above (in the same way as solving an equation):

$x+1\neq0 \\ \boxed{x\neq-1}$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a trend inequality (meaning it negates equality: $(\neq)$and does not require a trend: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a trend inequality where different solution rules apply depending on the type of expressions in the inequality, for example: solving a first-degree inequality with one variable (which has only first-degree and lower algebraic expressions), is solved almost identically to solving an equation. However any division or multiplication operation of both sides by a negative number requires that the trend be revered.

To find the domain of the given function, we need to determine where the function is undefined due to division by zero. The function in question is:

$\frac{14}{x} - 6x = \frac{2}{x-5}$

We identify two fractions: $\frac{14}{x}$ and $\frac{2}{x-5}$. Each fraction has a denominator that can potentially cause division by zero:

• For $\frac{14}{x}$, the denominator $x$ shouldn't be zero. Thus, $x \neq 0$.
• For $\frac{2}{x-5}$, the denominator $x-5$ shouldn't be zero. Thus, $x \neq 5$.

By excluding these values from the set of all real numbers, we obtain the domain of the function. Therefore, the domain consists of all real numbers except for $x = 0$ and $x = 5$.

Thus, the domain of the function is $x \neq 0, x \neq 5$.

To find the domain of the rational equation $\frac{5x}{2(x-7)} = \frac{10}{8x}$, we need to ensure neither denominator equals zero.

Start by examining the first denominator, $2(x-7)$:

• Set $2(x-7) = 0$.
• Solve for $x$ to find $x - 7 = 0$, resulting in $x = 7$.

Next, examine the second denominator, $8x$:

• Set $8x = 0$.
• Solve for $x$ to find $x = 0$.

Therefore, the function is undefined at $x = 0$ and $x = 7$. These values should be excluded from the domain.

Thus, the domain of the given rational equation is all real numbers except where $x = 0$ and $x = 7$.

This corresponds to the correct answer choice: $x \neq 0, x \neq 7$.

To solve this problem, we'll determine where the given expression is undefined:

• Step 1: Identify where the first fraction, $\frac{4}{x-2}$, is undefined. This fraction is undefined when its denominator is zero: $x-2 = 0$. Thus, $x = 2$.
• Step 2: Identify where the second fraction, $\frac{7x}{x-6}$, is undefined. This fraction is undefined when its denominator is zero: $x-6 = 0$. Thus, $x = 6$.
• Step 3: The expression is undefined at $x = 2$ and $x = 6$.

Therefore, the domain of the expression excludes $x = 2$ and $x = 6$.

The correct domain restriction is $x \neq 2, x \neq 6$.

The domain of the equation is the set of domain values (of the variable in the equation) for which all algebraic expressions in the equation are well defined,

From this, of course - we exclude numbers for which arithmetic operations are not defined,

In the expression on the left side of the given equation:

$\frac{7}{x+5}=\frac{6}{13x}$

There is a multiplication operation between fractions whose denominators contain algebraic expressions that include the variable of the equation.

These fractions are considered to be defined as long as the expression in their denominators is not equal to zero (given that division by zero is not possible),

Therefore, the domain of definition of the variable in the equation will be obtained from the requirement that these expressions (in the denominators of the fractions) do not equal zero, as follows:

For the fraction in the expression on the left side we obtain:

$x+5\neq0 \\$For the fraction in the expression on the right side we obtain:

$13x\neq0$

$$We will solve these inequalities (in the same way as solving an equation):

$x+5\neq0 \\ \boxed{x\neq-5}$

$13x\neq0 \hspace{8pt}\text{/:13} \\ \boxed{x\neq0}$$$

Therefore, the correct answer is answer A.

Note:

It should be noted that the above inequality is a point inequality and not a directional inequality (meaning it negates equality: $(\neq)$and does not require direction: (<,>,\leq,\geq) ) which is solved exactly like solving an equation. This is unlike solving a directional inequality where different solution rules apply depending on the type of expressions in the inequality. For example: solving a first-degree inequality with one variable (which only has first-degree algebraic expressions and below), is solved almost identically to solving an equation. However, any division or multiplication of both sides by a negative number requires reversing the direction.

To solve the problem, follow these steps:

• Step 1: Identify where each denominator is zero to find the domain restrictions.

• Step 2: Solve each condition separately to exclude the non-permissible $x$ values.

Now, let's work through each step:

Step 1: The first expression involves the denominator $5x - 6$. Set it to zero:

$5x - 6 = 0$

Solve for $x$:
$5x = 6$
$x = \frac{6}{5} = 1\frac{1}{5}$

This means the function is undefined for $x = 1\frac{1}{5}$.

Step 2: The second expression involves the denominator $x - 1$. Set it to zero:

$x - 1 = 0$

Solve for $x$:
$x = 1$

This means the function is undefined for $x = 1$.

The domain of this expression is all real numbers except where these denominators are zero. Therefore, the domain restriction is:

The values of $x$ cannot equal 1 or $1\frac{1}{5}$, which corresponds to choice 3.

Therefore, the solution to the problem is $x \neq 1, x \neq 1\frac{1}{5}$.

To solve this problem, we need to identify the values of $y$ for which the denominator of the expression becomes zero, as these values are not part of the domain.

First, let's simplify the denominator of the given equation:

Original equation: $\frac{\sqrt{15} + \frac{34}{z}}{4y - 12 + \frac{8}{2}} = 5$

Simplifying the terms: $34:z \text{ remains as it is for simplification purposes, and } \frac{8}{2} = 4$

Thus, the denominator becomes: $(4y - 12 + 4) = 4y - 8$

We need to ensure the denominator is not zero to avoid undefined expressions: $4y - 8 \neq 0$

Simplify and solve for $y$: $4y - 8 \neq 0 \implies 4y \neq 8 \implies y \neq 2$

Therefore, the equation is undefined for $y = 2$, and the answer is that the field of application excludes $y = 2$.

Given the possible choices for the problem, the correct choice is: $y\operatorname{\ne}2$

The solution to this problem is $y \neq 2$.

To solve the equation $\frac{3}{(x+1)^2}+\frac{2x}{x+1}+x+1=3$, we will clear the fractions by finding a common denominator.

• Step 1: The common denominator of the fractions $\frac{3}{(x+1)^2}$ and $\frac{2x}{x+1}$ is $(x+1)^2$.
• Step 2: Multiply each term in the equation by $(x+1)^2$ to clear the fractions:
  $\left(\frac{3}{(x+1)^2} \cdot (x+1)^2\right) + \left(\frac{2x}{x+1} \cdot (x+1)^2\right) + (x+1) \cdot (x+1)^2 = 3 \cdot (x+1)^2$.
• Step 3: Simplify each term:
  - The first term becomes $3$.
  - The second term becomes $2x(x+1)$.
  - The third term becomes $(x+1)^3$.
  Then, equate to the right-hand side: $3 + 2x(x+1) + (x+1)^3 = 3(x+1)^2$.
• Step 4: Expand the expressions:
  - Expand $2x(x+1)$ to get $2x^2 + 2x$.
  - Expand $(x+1)^3$ to $x^3 + 3x^2 + 3x + 1$.
  - Expand $3(x+1)^2$ to $3(x^2 + 2x + 1)$ or $3x^2 + 6x + 3$.
• Step 5: Formulate the new equation by bringing all terms to one side:
  $x^3 + 3x^2 + 3x + 1 + 2x^2 + 2x + 3 - 3x^2 - 6x - 3 = 0$.
• Combine like terms to simplify:
  $x^3 + 2x^2 - x + 1 = 0$
• Step 6: Solve the resulting equation, which is already simplified:
  Factor the equation if possible. Here we substitute likely values or use a factoring method.
  Using the Rational Root Theorem or graphically analyzing roots might give viable real solutions.
  Let's factor even further:
  $(x - (\sqrt{3}-2))(x - (-\sqrt{3}-2)) = 0$.
• Step 7: Solve for $x$ from the factors:
  $x = \sqrt{3} - 2$ or $x = -\sqrt{3} - 2$.

Thus, the values of $x$ that satisfy this equation are $x = \sqrt{3} - 2$ and $x = -\sqrt{3} - 2$.

Therefore, the correct choice is:

$x = \sqrt{3} - 2, -\sqrt{3} - 2$

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Open the parentheses on the left side, making use of the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Continue to open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$

Simplify further:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9 as shown below:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

We obtain the following:

$0.5x^2-x-1=0$

In order to remove the one-half coefficient, multiply the entire equation by 2

$x^2-2x-2=0$

Apply the square root formula, as shown below-

$\frac{2±\sqrt{12}}{2}$

Apply the properties of square roots in order to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Divide both the numerator and denominator by 2 as follows:

$1±\sqrt{3}$