Solve 1/(x - 2) 1/(x - 1) > 1/x

Question

Solve the inequality

Solve the inequality by testing the values in the interval

Solve the inequality by separating into cases

x \in (- \infty, 0) \cup (- 2 + 2, 1) \cup (2, 2 + 2)

Evaluate

\frac{1}{x - 2} \times \frac{1}{x - 1} > \frac{1}{x}

Find the domain

More Steps

\frac{1}{x - 2} \times \frac{1}{x - 1} > \frac{1}{x}, x \in (- \infty, 0) \cup (0, 1) \cup (1, 2) \cup (2, + \infty)

Multiply the terms

\frac{1}{( x - 2 ) ( x - 1 )} > \frac{1}{x}

Move the expression to the left side

\frac{1}{( x - 2 ) ( x - 1 )} - \frac{1}{x} > 0

Subtract the terms

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\frac{4 x - x ^{2} - 2}{( x - 2 ) ( x - 1 ) x} > 0

Set the numerator and denominator of \frac{4 x - x ^{2} - 2}{( x - 2 ) ( x - 1 ) x} equal to 0 to find the values of x where sign changes may occur

4 x - x^{2} - 2 = 0 (x - 2) (x - 1) x = 0

Calculate

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x = 2 + 2 x = - 2 + 2 (x - 2) (x - 1) x = 0

Calculate

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x = 2 + 2 x = - 2 + 2 x = 2 x = 1 x = 0

Determine the test intervals using the critical values

x < 0 0 < x < - 2 + 2 - 2 + 2 < x < 1 1 < x < 2 2 < x < 2 + 2 x > 2 + 2

Choose a value form each interval

x_{1} = - 1 x_{2} = \frac{- 2 + 2}{2} x_{3} = \frac{3 - 2}{2} x_{4} = \frac{3}{2} x_{5} = 3 x_{6} = 4

To determine if x < 0 is the solution to the inequality,test if the chosen value x = - 1 satisfies the initial inequality

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x < 0 is the solution x_{2} = \frac{- 2 + 2}{2} x_{3} = \frac{3 - 2}{2} x_{4} = \frac{3}{2} x_{5} = 3 x_{6} = 4

To determine if 0 < x < - 2 + 2 is the solution to the inequality,test if the chosen value x = \frac{- 2 + 2}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < - 2 + 2 is not a solution x_{3} = \frac{3 - 2}{2} x_{4} = \frac{3}{2} x_{5} = 3 x_{6} = 4

To determine if - 2 + 2 < x < 1 is the solution to the inequality,test if the chosen value x = \frac{3 - 2}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < - 2 + 2 is not a solution - 2 + 2 < x < 1 is the solution x_{4} = \frac{3}{2} x_{5} = 3 x_{6} = 4

To determine if 1 < x < 2 is the solution to the inequality,test if the chosen value x = \frac{3}{2} satisfies the initial inequality

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x < 0 is the solution 0 < x < - 2 + 2 is not a solution - 2 + 2 < x < 1 is the solution 1 < x < 2 is not a solution x_{5} = 3 x_{6} = 4

To determine if 2 < x < 2 + 2 is the solution to the inequality,test if the chosen value x = 3 satisfies the initial inequality

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x < 0 is the solution 0 < x < - 2 + 2 is not a solution - 2 + 2 < x < 1 is the solution 1 < x < 2 is not a solution 2 < x < 2 + 2 is the solution x_{6} = 4

To determine if x > 2 + 2 is the solution to the inequality,test if the chosen value x = 4 satisfies the initial inequality

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x < 0 is the solution 0 < x < - 2 + 2 is not a solution - 2 + 2 < x < 1 is the solution 1 < x < 2 is not a solution 2 < x < 2 + 2 is the solution x > 2 + 2 is not a solution

The original inequality is a strict inequality,so does not include the critical value ,the final solution is x \in (- \infty, 0) \cup (- 2 + 2, 1) \cup (2, 2 + 2)

x \in (- \infty, 0) \cup (- 2 + 2, 1) \cup (2, 2 + 2)

Check if the solution is in the defined range

x \in (- \infty, 0) \cup (- 2 + 2, 1) \cup (2, 2 + 2), x \in (- \infty, 0) \cup (0, 1) \cup (1, 2) \cup (2, + \infty)

Solution

x \in (- \infty, 0) \cup (- 2 + 2, 1) \cup (2, 2 + 2)

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