26.
-578-V-20x® + V-45.18, where x is a positive real number.
the following:
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Answer:
We have to prove that, [
x
b
x
a
]
a+b−c
[
x
c
x
b
]
b+c−a
[
x
a
x
c
]
c+a−b
=1
Let a=[
x
b
x
a
]
a+b−c
[
x
c
x
b
]
b+c−a
[
x
a
x
c
]
c+a−b
=(x
a−b
)
a+b−c
(x
b−c
)
b+c−a
(x
c−a
)
c+a−b
=x
(a−b)(a+b−c)
×x
(b−c)(b+c−a)
×x
(c−a)(c+a−b)
=x
(a−b)(a+b)−c(a−b)
×x
(b−c)(b+c)−a(b−c)
×x
(c−a)(c+a)−b(c−a)
=x
a
2
−b
2
−ca+bc
×x
b
2
−c
2
−ab+ac
×x
c
2
−a
2
−bc+ba
a=x
0
=1
∴[
x
b
x
a
]
a+b−c
[
x
c
x
b
]
b+c−a
[
x
a
x
c
]
c+a−b
=1
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