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How to use it?
- Inequation:
- 2*(|x|)-1<13
- 7^(-|x-3|)*log(2,6*x-(x^2)+7)>=1
- 5^x^2*16^x-1≥5
- (8-x)/(x-10)≤2/2-x
- Sum of series:
-
-1
- Derivative of:
-
-1
- Integral of d{x}:
-
-1
-
Identical expressions
- |x- one |-(six /(|x- one |))<- one
- module of x minus 1| minus (6 divide by (|x minus 1|)) less than minus 1
- module of x minus one | minus (six divide by (|x minus one |)) less than minus one
- |x-1|-6/|x-1|<-1
- |x-1|-(6 divide by (|x-1|))<-1
-
Similar expressions
- |x+1|-(6/(|x-1|))<-1
- |x-1|-(6/|x-1|)<-5
- |x-1|-(6/(|x+1|))<-1
- |x-1|-(6/(|x-1|))<+1
- |x-1|+(6/(|x-1|))<-1
|x-1|-(6/(|x-1|))<-1 inequation
The solution
You have entered
[src]
6
|x - 1| - ------- < -1
|x - 1|
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} < -1$$
|x - 1| - 6/|x - 1| < -1
Detail solution
Given the inequality:
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} < -1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} = -1$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 3$$
$$x_{1} = -1$$
$$x_{2} = 3$$
This roots
$$x_{1} = -1$$
$$x_{2} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$-1.1$$
substitute to the expression
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} < -1$$
$$- \frac{6}{\left|{-1.1 - 1}\right|} + \left|{-1.1 - 1}\right| < -1$$
but
Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < 3$$
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} < -1$$
To solve this inequality, we must first solve the corresponding equation:
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} = -1$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 3$$
$$x_{1} = -1$$
$$x_{2} = 3$$
This roots
$$x_{1} = -1$$
$$x_{2} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$-1.1$$
substitute to the expression
$$\left|{x - 1}\right| - \frac{6}{\left|{x - 1}\right|} < -1$$
$$- \frac{6}{\left|{-1.1 - 1}\right|} + \left|{-1.1 - 1}\right| < -1$$
-0.757142857142857 < -1
but
-0.757142857142857 > -1
Then
$$x < -1$$
no execute
one of the solutions of our inequality is:
$$x > -1 \wedge x < 3$$
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x1 x2
Solving inequality on a graph
Rapid solution
[src]
Or(And(-1 < x, x < 1), And(1 < x, x < 3))
$$\left(-1 < x \wedge x < 1\right) \vee \left(1 < x \wedge x < 3\right)$$
((-1 < x)∧(x < 1))∨((1 < x)∧(x < 3))
Rapid solution 2
[src]
(-1, 1) U (1, 3)
$$x\ in\ \left(-1, 1\right) \cup \left(1, 3\right)$$
x in Union(Interval.open(-1, 1), Interval.open(1, 3))