simplify {(√a+x)-(√a-x)} ÷ {(√a+x) +(a-x) }
Answers
Step-by-step explanation:
=
[
a
+x]−[
a
−x]
[
a
+x]+[
a
−x]
= \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]-[ \sqrt{a}-x]} * \frac{[ \sqrt{a}+x]+[ \sqrt{a}-x]}{[ \sqrt{a}+x]+[ \sqrt{a}-x]}=
[
a
+x]−[
a
−x]
[
a
+x]+[
a
−x]
∗
[
a
+x]+[
a
−x]
[
a
+x]+[
a
−x]
= \frac{ [ \sqrt{a}+x]+[ \sqrt{a}-x]^{2} }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }=
[
a
+x]
2
−[
a
−x]
2
[
a
+x]+[
a
−x]
2
= \frac{ [ \sqrt{a}+x]^{2}+[ \sqrt{a}-x]^{2}+2[ \sqrt{a}+x][ \sqrt{a}-x] }{[ \sqrt{a}+x]^{2} -[ \sqrt{a}-x]^{2} }=
[
a
+x]
2
−[
a
−x]
2
[
a
+x]
2
+[
a
−x]
2
+2[
a
+x][
a
−x]
= \frac{4a}{4x \sqrt{a}}=
4x
a
4a
= \frac{ \sqrt{a} }{x}=
x
a
Now if
x= \frac{2ab}{1+ b^{2} }x=
1+b
2
2ab
then we get
= \frac{ \sqrt{a} }{2ab}*[1+ b^{2}]=
2ab
a
∗[1+b
2
]
= \frac{ [1+ b^{2}] }{2b \sqrt{a} }=
2b
a
[1+b
2
]