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| 1/x - 2/(x+1) |  =  2/x

Because the left-hand side is contained within absolute value bars, its value must be positive (or zero).

Thus, the right-hand side must also be positive; for  2/x  to be positive,  x  must be positive.

Simplifying the left-hand side:  | 1/x - 2/(x + 1) |  =  | 1(x + 1)/[x(x + 1)] - 2x/[x(x + 1)] |  

                                                                          =  | (x + 1 - 2x) / [ x(x + 1) ] |

                                                                          =  | (1 - x) / [ x(x + 1) ] |

Therefore:  | (1 - x) / [ x(x + 1) ] |  =  2/x

Since it is an absolute value equation, there are two possibilities:

     either     (1 - x) / [ x(x + 1) ]  =  2/x     or      (1 - x) / [ x(x + 1) ]  =  -2/x

If       (1 - x) / [ x(x + 1) ]  =  2/x

--->   (1 - x) / [ x(x + 1) ]  =  2(x + 1) / [ x(x + 1) ]

--->   1 - x  =  2(x + 1)

--->   1 - x  =  2x + 2

--->   -1  =  3x

--->   x  =  -1/3          <--- But this is impossible, because  x  must be positive.

If      (1 - x) / [ x(x + 1) ]  =  -2/x

--->  (1 - x) / [ x(x + 1) ]  =  -2(x + 1) / [x(x + 1) ]

--->   1 - x  =  -2(x + 1)

--->   1 - x  =  -2x - 2

--->   3  =  -x

--->   x  =  -1/3          <--- But this is also impossible.

So, I think that the correct answer is  d) it doesn't have any real roots.

Solve for x:
1/x-2/(x+1) = 2/x

Multiply both sides by x:
1-(2 x)/(x+1) = 2

Bring 1-(2 x)/(x+1) together using the common denominator x+1:
(1-x)/(x+1) = 2

Multiply both sides by x+1:
1-x = 2 (x+1)

Expand out terms of the right hand side:
1-x = 2 x+2

Subtract 2 x+1 from both sides:
-3 x = 1

Divide both sides by -3:
Answer: |  x = -1/3        Looks like the answer is "C"

Hallo Guest!

An equation

How many real roots does it have:

$|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}$

a)Exactly 3 real roots

b)Exactly 2 real roots

c)Exactly 1 real root

d)It doesn't have any real roots

$|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}$

$\frac{x+1-2x}{x²+x}= \frac{2x+2}{x²+x}$                                  [ * ( x² + x)

x + 1 - 2x = 2x + 2

- 3x = 1

x = - 1 / 3

Sample:

1 / (- 1 / 3) - 2 / (1 - 1 / 3) = 2 / (- 1 / 3)

       - 3      -              3       =   - 6

                                     - 6 = - 6

c)Exactly 1 real root

Greeting asinus :- )

   !

We can see by plotting a graph that geno is correct:

Hallo geno and Alan!

An equation

How many real roots does it have:

$|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x}$

a)Exactly 3 real roots

b)Exactly 2 real roots

c)Exactly 1 real root

d)It doesn't have any real roots

Of course you are right.
The condition "abs" matters:
Without "absolutely" is x = - 1/3,
with the "absolutely" no root.

It applies
d) It does not have any real roots

Greeting asinus :- )

   !

An equation

$|\frac{1}{x}-\frac{2}{x+1}|=\frac{2}{x} \qquad \qquad x\ne 0$

How many real roots does it have:

a)Exactly 3 real roots

b)Exactly 2 real roots

c)Exactly 1 real root

d)It doesn't have any real roots

$\begin{array}{rcll} |\frac{1}{x}-\frac{2}{x+1}| &=& \frac{2}{x} \qquad & | \qquad \text{square both sides} \\ \left(\frac{1}{x}-\frac{2}{x+1} \right)^2 &=& \left(\frac{2}{x} \right)^2 \\ \left(\frac{x+1-2x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \left(\frac{ 1- x}{x(x+1)} \right)^2 &=& \frac{4}{x^2} \\ \frac{ (1- x)^2}{x^2(x+1)^2} &=& \frac{4}{x^2} \qquad & | \qquad \cdot x^2 \qquad \text{solution }~ x = 0 ~ \text{impossible}\\ \frac{ (1- x)^2}{ (x+1)^2} &=& 4 \\ 4\cdot(x+1)^2 &=& (1- x)^2 \\ 4\cdot( x^2+2x+1) &=& (1-2x+x^2) \\ 4x^2 + 8x+4 &=& 1-2x+x^2 \\ 3x^2 + 10x+3 &=& 0\\ \end{array}$

$\begin{array}{|rcll|} \hline 3x^2 + 10x+3 &=& 0 \qquad \qquad a = 3 \qquad b=10 \qquad c = 3 \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{10^2-4\cdot 3 \cdot 3} } { 2\cdot 3 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{100-36} } { 6 } \\\\ x_{1,2} &=& \frac{ -10 \pm \sqrt{64} } {6 } \\\\ x_{1,2} &=& \frac{ -10 \pm 8 } {6 } \\\\ x_{1} &=& \frac{ -10 + 8 } { 6 } \\ x_{1} &=& \frac{ -2 } { 6 } \\ \mathbf{ x_{1} }&\mathbf{=}& \mathbf{ - \frac13 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ x_{2} &=& \frac{ -10 - 8 } { 6 } \\ x_{2} &=& \frac{ -18 } { 6 } \\ \mathbf{ x_{2} }&\mathbf{=}& \mathbf{ - 3 } \qquad &| \qquad \text{solution impossible }~ x > 0 \\\\ \hline \end{array}$

d) It doesn't have any real roots