Question

$\frac{1}{ \sqrt{2} - 2 } + \frac{1}{ \sqrt{2} + 2 } + \sqrt{2} + 2$
please with steps​

Answer 1

Answer:

6+$\sqrt{2}$

Step-by-step explanation:

-1+1+($\sqrt{2}$)(2-$\sqrt{2}$)(2+$\sqrt{2}$) + (2)(2-$\sqrt{2}$)(2+$\sqrt{2}$)/(2-$\sqrt{2}$)(2+$\sqrt{2}$)

(2$\sqrt{2}$-2)(2+$\sqrt{2}$) +(4-2$\sqrt{2}$)(2+$\sqrt{2}$)/4-2

4$\sqrt{2}$+4-4-2$\sqrt{2}$+8+4$\sqrt{2}$-4$\sqrt{2}$+4/2

4$\sqrt{2}$-2$\sqrt{2}$+8+4/2

2$\sqrt{2}$+12/2

2($\sqrt{2}$+6)/2

6+$\sqrt{2}$

Answer 2

Answer:

$1$

Step-by-step explanation:

Let's do addition of first two terms

LCM of first two terms: ($\sqrt{2} -2$)($\sqrt{2} +2$)

Therefore equation would be:

$(\sqrt{2} -2) + (\sqrt{2} +2) / (\sqrt{2} -2)* (\sqrt{2} +2)$

= $\sqrt2 + \sqrt2 + 2 -2 / (\sqrt2)^{2} - 2^{2}$ As $(a-b)(a+b) = a^{2} - b^{2}$

= $2\sqrt{2} / 2-4$

= $(2\sqrt{2} /(-2))$

So now we got the value of first 2 terms i.e : $(2\sqrt{2} /(-2))$

Therefore:

=  $(2\sqrt{2} /(-2)) + \sqrt{2} + 2$

LCM: -2

= $((2\sqrt{2}) + (-2 * \sqrt{2}) + -2 * 2) / (-2)$

= $(2\sqrt{2} - 2\sqrt{2} -2) / (-2)$

= $0-2 /-2$

= $-2 / -2$

= $1$

Please mark me brainliest, I typed for half an hour