Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation ((1/5))^(x)^(2)*5^(2*x+2)=(1/5)^(6)
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Equation x^3-x^2-x-1=0
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Equation -x=3
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Equation x-6/2-x/3=3
- Express {x} in terms of y where:
- 9*x+10*y=5
- -4*x+19*y=-13
- -5*x-6*y=-1
- -7*x+15*y=18
- Graphing y =:
- x/(x-1)^2
- Integral of d{x}:
- x/(x-1)^2
- Derivative of:
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x/(x-1)^2
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Identical expressions
- x/(x- one)^ two = zero
- x divide by (x minus 1) squared equally 0
- x divide by (x minus one) to the power of two equally zero
- x/(x-1)2=0
- x/x-12=0
- x/(x-1)²=0
- x/(x-1) to the power of 2=0
- x/x-1^2=0
- x/(x-1)^2=O
- x divide by (x-1)^2=0
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Similar expressions
- x/(x+1)^2=0
x/(x-1)^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\frac{x}{\left(x - 1\right)^{2}} = 0$$
the denominator
$$x - 1$$
then
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x = 0$$
solve the resulting equation:
1.
$$x = 0$$
We get the answer: x1 = 0
but
The final answer:
$$x_{1} = 0$$
$$\frac{x}{\left(x - 1\right)^{2}} = 0$$
the denominator
$$x - 1$$
then
x is not equal to 1
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x = 0$$
solve the resulting equation:
1.
$$x = 0$$
We get the answer: x1 = 0
but
x is not equal to 1
The final answer:
$$x_{1} = 0$$
The graph