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-
How to use it?
- Inequation:
- 2*(|x|)-1<13
- (((3x+7)^0.5)+2)^(1-x)>=(3+x)^(1-x)
- 5^(x^2+x)>-1
- log(x^2-3п/2)(3-2^x)>0
- Sum of series:
-
13
- Derivative of:
-
13
- Canonical form:
- 13
-
Identical expressions
- two *(|x|)- one < thirteen
- 2 multiply by ( module of x|) minus 1 less than 13
- two multiply by ( module of x|) minus one less than thirteen
- 2(|x|)-1<13
- 2|x|-1<13
-
Similar expressions
- abs(2*|x|-1)<=x+2
- 2*(|x|)+1<13
- (x^2+x+1)^((x-10)/(x-3))<=(x^2+x+1)^3
- (35(2^|x|))/(4+10(2^|x|)-6(4^|x|))=>((2^|x|)+2)/(3(2^|x|)+1)+(3(2^|x|)-1)/(2-2^|x|)
- 35*2^|x|/(4+10*2^|x|-6*4^|x|)>=(2^|x|+2)/(3*2^|x|+1)+(3*2^|x|-1)/(2-2^|x|)
- (2^|x|-1)*(3-2x-x^2)/(|x-1|-6)>=0
2*(|x|)-1<13 inequation
The solution
Detail solution
Given the inequality:
$$2 \left|{x}\right| - 1 < 13$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \left|{x}\right| - 1 = 13$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x \geq 0$$
or
$$0 \leq x \wedge x < \infty$$
we get the equation
$$2 x - 14 = 0$$
after simplifying we get
$$2 x - 14 = 0$$
the solution in this interval:
$$x_{1} = 7$$
2.
$$x < 0$$
or
$$-\infty < x \wedge x < 0$$
we get the equation
$$2 \left(- x\right) - 14 = 0$$
after simplifying we get
$$- 2 x - 14 = 0$$
the solution in this interval:
$$x_{2} = -7$$
$$x_{1} = 7$$
$$x_{2} = -7$$
$$x_{1} = 7$$
$$x_{2} = -7$$
This roots
$$x_{2} = -7$$
$$x_{1} = 7$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-7 + - \frac{1}{10}$$
=
$$- \frac{71}{10}$$
substitute to the expression
$$2 \left|{x}\right| - 1 < 13$$
$$-1 + 2 \left|{- \frac{71}{10}}\right| < 13$$
but
Then
$$x < -7$$
no execute
one of the solutions of our inequality is:
$$x > -7 \wedge x < 7$$
$$2 \left|{x}\right| - 1 < 13$$
To solve this inequality, we must first solve the corresponding equation:
$$2 \left|{x}\right| - 1 = 13$$
Solve:
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x \geq 0$$
or
$$0 \leq x \wedge x < \infty$$
we get the equation
$$2 x - 14 = 0$$
after simplifying we get
$$2 x - 14 = 0$$
the solution in this interval:
$$x_{1} = 7$$
2.
$$x < 0$$
or
$$-\infty < x \wedge x < 0$$
we get the equation
$$2 \left(- x\right) - 14 = 0$$
after simplifying we get
$$- 2 x - 14 = 0$$
the solution in this interval:
$$x_{2} = -7$$
$$x_{1} = 7$$
$$x_{2} = -7$$
$$x_{1} = 7$$
$$x_{2} = -7$$
This roots
$$x_{2} = -7$$
$$x_{1} = 7$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$-7 + - \frac{1}{10}$$
=
$$- \frac{71}{10}$$
substitute to the expression
$$2 \left|{x}\right| - 1 < 13$$
$$-1 + 2 \left|{- \frac{71}{10}}\right| < 13$$
66/5 < 13
but
66/5 > 13
Then
$$x < -7$$
no execute
one of the solutions of our inequality is:
$$x > -7 \wedge x < 7$$
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