91Ó°ÊÓ

An equation is a mathematical statement that asserts the equality of two expressions. Essentially, it shows that two things are equal. Solving an equation means finding all the values of the variable that make the equation true.
In the exercise problem, you are given the equation \( \frac{x(x-2)}{(x-1)^2} = 0 \). To find the solutions, you need to set the numerator equal to zero since a fraction equals zero only when its numerator is zero, assuming the denominator is not zero. Thus, we solve \( x(x-2) = 0 \) which gives us the solutions \( x = 0 \) and \( x = 2 \). It is crucial to check that these solutions do not make the denominator zero. In this case, \( x = 1 \) makes the denominator zero and is hence not a solution. When solving equations, always ensure the denominator is not zero at the solution points, as this makes the equation undefined. Check each solution to ensure it is part of the domain of the original equation.

Inequalities are statements about the relative size of two expressions. They are similar to equations, but instead of asserting that two expressions are equal, they determine the relationship between them through signs like greater than \(>\), less than \(<\), greater than or equal to \( \geq \), and less than or equal to \( \leq \).
In the inequality problem \( \frac{x(x-2)}{(x-1)^2} > 0 \), you need to find intervals where the expression is positive. You define intervals around critical points, solve each to see the sign of the interval, and write the solution as a union of intervals where the inequality holds true.
For the exercise, the critical points are \( x = 0 \), \( x = 1 \), and \( x = 2 \). Choose test points from intervals to determine the sign of the expression in each interval:

• \( x < 0 \)
• \( 0 < x < 1 \)
• \( 1 < x < 2 \)
• \( x > 2 \)

Select one point from each interval to plug into the inequality, and determine whether it is positive or negative. The expression is positive in the intervals \((- \infty, 0)\) and \((2, +\infty)\). These contribute to the solution of the inequality.

Expressions in algebra are combinations of numbers, variables, and operators (like +, -, *, and /) that represent a particular value. Unlike equations, expressions do not have equals signs indicating a relationship.
The expression in the given problem, \( \frac{(x-1)(2 x)- left(x^{2} ight)(1)}{(x-1)^{2}} \), needs to be simplified to a more usable form. Simplifying involves combining like terms and employing the distributive property.
First, expand \((x-1)(2x)\) to get \(2x^2 - 2x\). Then subtract \(x^2\) to get the simplified numerator \(x^2 - 2x\). So now, the expression becomes \( \frac{x(x-2)}{(x-1)^2} \).
Understanding expressions and their simplification is key to solving equations and inequalities. It allows easier manipulation and clearer insight into the problem structure, which is essential for finding solutions.

Critical points are specific values in a function where the behavior changes, including values where the function may be undefined or zero. They are essential for solving inequalities because they partition the real number line into intervals, where the function's behavior can be tested.
In the context of the given exercises, the critical points arise from setting the numerator \(x(x-2)\) equal to zero and the denominator \((x-1)^2\) equal to zero:

• \(x = 0 \)
• \(x = 2 \)
• \(x = 1 \)

Each point divides the real number line into distinct parts, and analyzing the function in each of these intervals gives insight into where the inequality holds.
While \(x = 0\) and \(x = 2\) are solutions for the equation, for the inequality solution \((x-1)^2\) = 0 (at \(x = 1\)) identifies undefined points. For the inequality \(\frac{x(x-2)}{(x-1)^2} > 0\), critical points help define intervals to evaluate sign changes. Test values in each interval ensure accurate representation of where the inequality holds.