Question

Solve for $\rm{x:}$

$\sf{\dfrac{x-1}{2x+1}+\dfrac{2x+1}{x-1} = 2}$

• Note that x is not equal to 1 and $\bf{\dfrac{-1}{2}}$​

Answer 1

Step-by-step explanation:

x−12x+1+2x+1x−1=2

(x−1)2+(2x+1)2(x−1)(2x+1)=2

x2−2x+1+4x2+4x+1=2(x−1)(2x+1)

5x2+2x+2=4x2−2x−2

x2+4x+4=0

(x+2)2=0

x=−2,−2

Answer 2

$\large {\pmb{☆ \: Given :}}$

$\tt \frac{x - 1}{2x + 1} + \frac{2x + 1}{x - 1} = 2 \: ; \: x≠1, \: \frac{-1}{2}$

$\large {\pmb{☆ \: To \: find:}}$

We have to find the value of x.

$\large {\pmb{☆ \: Solution:}}$

• Refer the attachments.

$\large {\pmb{☆ \: Know \: more:}}$

• $\tt a^3-b^3=(a-b)^3 +3ab(a-b)$

• $\tt a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

• $\tt (a-b)(a+b)=a^2-b^2$

• $\tt (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

• $\tt (a-b-c)^2=a^2+b^2+c^2-2ab+2bc-2ca$

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It shows that my answer contains rude words...So I've attached the pics!

Kindly adjust with the handwriting -,-