Advertisements
Advertisements
प्रश्न
If `x = (6ab)/(a + b)`, find the value of `(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b)`.
उत्तर
`x = (6ab)/(a + b)`
`=> x/(3a) = (2b)/(a + b)`
Aplying compinendo and dividend,
`(x + 3a)/(x - 3a) = (2b + a + b)/(2b - a - b)`
`(x + 3a)/(x - 3a) = (3b + a)/(b - a)` ...(1)
Again, `x = (6ab)/(a + b)`
`=> x/(3b) = (2a)/(a + b)`
Applying componendo and dividendo,
`(x + 3b)/(x - 3b) = (2a + a + b)/(2a - a - b)`
`(x + 3b)/(x - 3b) = (3a + b)/(a - b)` ...(2)
From (1) and (2)
`(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b) = (3b + a)/(b -a) + (3a + b)/(a - b)`
`(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b) = (-3b -a + 3a + b)/(a - b)`
`(x + 3a)/(x - 3a) + (x + 3b)/(x - 3b) = (2a - 2b)/(a - b) = 2`
APPEARS IN
संबंधित प्रश्न
Given `(x^3 + 12x)/(6x^2 + 8) = (y^3+ 27y)/(9y^2 + 27)`. Using componendo and dividendo find x : y.
Given, `a/b = c/d`, prove that: `(3a - 5b)/(3a + 5b) = (3c - 5d)/(3c + 5d)`
If `(a - 2b - 3c + 4d)/(a + 2b - 3c - 4d) = (a - 2b + 3c - 4d)/(a + 2b + 3c + 4d)`, show that: 2ad = 3bc.
If (4a + 9b)(4c – 9d) = (4a – 9b)(4c + 9d), prove that: a : b = c : d.
If a : b = c : d , then prove that `("a"^2 + "ab" +
"b"^2)/("a"^2 - "ab" + "b"^2) = ("c"^2 + "cd"+ "d"^2)/("c"^2 - "cd" + "d"^2)`
If a : b :: c : d :: e : f, then prove that `("ae" + "bf")/("ae" - "bf")` = `("ce" + "df")/("ce" - "df")`
If `a/b = c/d,` show that (9a + 13b) (9c - 13d) = (9c + 13b) (9a - 13d).
If x = `(2a + b)/(a + b)` find the value of `(x + a)/(x - a) + (x + b)/(x - b)`
Find x from the following equations : `(sqrt(1 + x) + sqrt(1 - x))/(sqrt(1 + x) - sqrt(1 - x)) = a/b`
If (m + n) : (n – m) = 5 : 2 ; m : n is ______.
