Advertisements
Advertisements
प्रश्न
If ax = by = cz and abc = 1, show that
`(1)/x + (1)/y + (1)/z` = 0.
उत्तर
ax = by = cz
So, ax = by ⇒ a =`"b"^(y/x) .....("Using" "a"^(1/"n") = root("n")("a"))`
by = cz ⇒ c = `"b"^(y/z) .....("Using" "a"^(1/"n") = root("n")("a"))`
and abc = 1
⇒ `"b"^(y/x) · "b"·"b"^(y/z)` = 1
⇒ `"b"^(y/x) · "b"·"b"^(y/z)` = 1
⇒ `"b"^(y/x + 1 + y/z)` = b° ......(Using a° = 1)
⇒ `y/x + 1 + y/z` = 0
Divide throughout by y.
⇒ `(1)/x + (1)/y + (1)/z` = 0
Hence proved.
APPEARS IN
संबंधित प्रश्न
Find x, if : `sqrt( 2^( x + 3 )) = 16`
Find x, if : `(root(3)( 2/3))^( x - 1 ) = 27/8`
Solve for x : 9x+2 = 720 + 9x
Solve for x:
22x+1= 8
Solve for x:
`"p"^-5 = (1)/"p"^(x + 1)`
Find the value of k in each of the following:
`(root(3)(8))^((-1)/(2)` = 2k
If a = `2^(1/3) - 2^((-1)/3)`, prove that 2a3 + 6a = 3
If x = `3^(2/3) + 3^(1/3)`, prove that x3 - 9x - 12 = 0
Show that : `(1)/(1 + "a"^("p"- "q")) + (1)/(1 + "a"^("q"- "p")`
If 2x = 3y = 12z ; show that `(1)/z = (1)/y + (2)/x`.