If 2^x=3^y=6^-z then 1/x+1/y+1/z=?
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Given, 2x = 3y = 6-z
2x = 6-z
⇒ 2 = 6-z/x -------------(i) and
3y = 6-z
⇒ 3 = 6-z/y -------------(ii)
Multiplying (i) and (ii) equations
⇒ 2×3 = (6-z/x) (6-z/y)
⇒ 6 = 6*(-z/x + -z/y)
⇒ 6/6 = (-z/x) +(-z/y)
⇒ 1= -z(1/x +1/y)
⇒ 1/-z = 1/x +1/y
therefore, 1/x + 1/y + 1/z = 0
2x = 6-z
⇒ 2 = 6-z/x -------------(i) and
3y = 6-z
⇒ 3 = 6-z/y -------------(ii)
Multiplying (i) and (ii) equations
⇒ 2×3 = (6-z/x) (6-z/y)
⇒ 6 = 6*(-z/x + -z/y)
⇒ 6/6 = (-z/x) +(-z/y)
⇒ 1= -z(1/x +1/y)
⇒ 1/-z = 1/x +1/y
therefore, 1/x + 1/y + 1/z = 0
Anonymous:
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