Advertisements
Advertisements
प्रश्न
Give `(x^3 + 12x)/(6x^2 + 8) = (y^3 + 27y)/(9y^2 + 27)` Using componendo and dividendo find x : y.
उत्तर
Give `(x^3 + 12x)/(6x^2 + 8) = (y^3 + 27y)/(9y^2 + 27)`
Using componendo-dividendo, we have
`(x^3 + 12x + 6x^2 + 8)/(x^3 + 12x - 6x^2 - 8) = (y^3 + 27y + 9y^2 + 27)/(y^3 + 27y 9y^2 - 27)`
⇒ `(x + 2)^3/(x 2)^3 = (y + 3)^3/(9y - 3)^3`
⇒ `((x + 2)/(x - 2))^3 = ((y + 3)/(y - 3))^3`
⇒ `(x + 2)/(x - 2) = (y + 3)/(y - 3)`
Again using componendo-dividendo, we get
`(x + 2 + x - 2)/(x + 2 - x + 2) = (y + 3 + y - 3)/(y + 3 - y + 3)`
⇒ `(2x)/(4) = (2y)/(3)`
⇒ `x/(2) = y/(3)`
⇒ `x/y = (2)/(3)`
Thus the required ratio is x : y = 2 : 3.
APPEARS IN
संबंधित प्रश्न
Given x = `(sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) + sqrt(a^2 - b^2))`
Use componendo and dividendo to prove that b^2 = (2a^2x)/(x^2 + 1)
If `(x^2 + y^2)/(x^2 - y^2) = 2 1/8`, find: `x/y`
if `(3a + 4b)/(3c + 4d) = (3a - 4b)/(3c - 4d)` Prove that `a/b = c/d`.
If (4a + 5b) (4c – 5d) = (4a – 5d) (4c + 5d), prove that a, b, c, d are in proporton.
If (11a² + 13b²) (11c² – 13d²) = (11a² – 13b²)(11c² + 13d²), prove that a : b :: c : d.
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(2 - x) + sqrt(2 + x))/(sqrt(2 - x) - sqrt(2 + x)` = 3
Find x from the following equations : `(sqrt(a + x) + sqrt(a - x))/(sqrt(a + x) - sqrt(a - x)) = c/d`
Given `x = (sqrt(a^2 + b^2) + sqrt(a^2 - b^2))/(sqrt(a^2 + b^2) - sqrt(a^2 - b^2)`. Use componendo and dividendo to prove that: `b^2 = (2a^2x)/(x^2 + 1)`
If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of `(3x^4 + 5y^4)/(3x^4 - 5y^4)`