Question

hey friends please help me out with this x-2/x+2+x+2/x-2=4​

Answer 1

Answer:

Check the file I've attached the answer

Step-by-step explanation:

bc9bd6c4d3e889497d0fbd38cfc10d46.docx

Answer 2

ANSWER:

Given:

• (x - 2)/(x + 2) + (x + 2)/(x - 2) = 4

To Find:

• Value of x

Solution:

We are given that,

$:\implies\dfrac{x-2}{x+2}+\dfrac{x+2}{x-2}=4$

Now we'll take LCM,

$:\implies\dfrac{(x-2)(x-2)+(x+2)(x+2)}{(x+2)(x-2)}=4$

So,

$:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4$

We know that,

$:\hookrightarrow(a\pm b)^2=a^2\pm2ab+b^2$

And,

$:\hookrightarrow(a+b)(a-b)=a^2-b^2$

So,

$:\implies\dfrac{(x-2)^2+(x+2)^2}{(x+2)(x-2)}=4$

$:\implies\dfrac{(x^2-4x+4)+(x^2+4x+4)}{(x^2-4)}=4$

Opening brackets,

$:\implies\dfrac{x^2-4x+4+x^2+4x+4}{x^2-4}=4$

$:\implies\dfrac{2x^2+8}{x^2-4}=4$

Transposing x² - 4 to RHS,

$:\implies2\!\!\!/\:(x^2+4)=4\!\!\!/^2\:(x^2-4)$

So,

$:\implies x^2+4=2(x^2-4)$

Opening the brackets,

$:\implies x^2+4=2x^2-8$

Transposing LHS to RHS,

$:\implies 0=2x^2-8-(x^2+4)$

So,

$:\implies 0=2x^2-8-x^2-4$

$:\implies x^2-12=0$

Transposing 12 to RHS,

$:\implies x^2=12$

$:\implies x=\pm\sqrt{12}$

Therefore,

$:\implies\bf x=\pm2\sqrt2$

Hence, the value of x is ±2√2.