please find the answer fast
Answers
Step-by-step explanation:
\rm {\huge\star} \: Given⋆Given
\bigg(\rm\dfrac{1}{1 + x^{a - b}} \bigg) + \bigg( \dfrac{1}{1 + x ^{b - a} } \bigg) = 1(
1+x
a−b
1
)+(
1+x
b−a
1
)=1
In any number when you get a negative power it means you have to make it positive by twisting it or it means you have to divide. In this case , we have to divide. -B means ÷ B So, let's see!!
\rm \longmapsto \bigg( \dfrac{1}{1 + \dfrac{x ^{a} }{x ^{b} } } \bigg) + \bigg( \dfrac{1}{1 + \dfrac{x^{b} }{x ^{a} } } \bigg) = 1⟼(
1+
x
b
x
a
1
)+(
1+
x
a
x
b
1
)=1
Now , the value which is at down will be getting multiplied by 1 also which is at down will remain as denominator.
\rm \longmapsto \bigg( \dfrac{1}{ \dfrac{x ^{a} + x ^{b} }{x ^{b} } } \bigg) + \bigg( \dfrac{1}{ \dfrac{x^{b} + {x}^{a} }{x ^{a} } } \bigg) = 1⟼(
x
b
x
a
+x
b
1
)+(
x
a
x
b
+x
a
1
)=1
Now the denominator will go to up as soon as we do reciprocal of it .
So, it becomes
\rm{ \tt \leadsto} \dfrac{ {x}^{b} }{ {x}^{a} + {x}^{b} } + \dfrac{ {x}^{a} }{ {x} ^{b } + {x}^{a} } = 1⇝
x
a
+x
b
x
b
+
x
b
+x
a
x
a
=1
Now, the denominator are same because ; a+b = b+a
So, let's solve it!!
\rm{ \tt \leadsto} \cancel \dfrac{ {x}^{b} + {x}^{a} }{ {x}^{a} + {x}^{b} } \longmapsto1=1⇝
x
a
+x
b
x
b
+x
a
⟼1=1
\rm {\huge \therefore} \: On \: simplifying \: the \: value \: we \: got \: the \: answer \: \bf1∴Onsimplifyingthevaluewegottheanswer1
Hence proved
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Check the image for answer.
Glad to help you.
