Advertisements
Advertisements
प्रश्न
If a ≠ 0 and `a - 1/a` = 4 ; find : `( a^2 + 1/a^2 )`
उत्तर
`( a - 1/a )^2 = a^2 + 1/a^2 - 2`
⇒ `a^2 + 1/a^2 = ( a - 1/a )^2 + 2`
⇒ `a^2 + 1/a^2 = (4)^2 + 2 [ ∵ a - 1/a = 4]`
⇒ `a^2 + 1/a^2` = 18
APPEARS IN
संबंधित प्रश्न
Simplify.
(3r − 2k)3 + (3r + 2k)3
Find the cube of : 3a- 2b
Find the cube of : 5a + 3b
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
If `a^2 + 1/a^2` = 18; a ≠ 0 find :
(i) `a - 1/a`
(ii) `a^3 - 1/a^3`
Use property to evaluate : 73 + 33 + (-10)3
Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
If `5x + (1)/(5x) = 7`; find the value of `125x^3 + (1)/(125x^3)`.
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"a"^3 - (1)/"a"^3`
Expand: (3x + 4y)3.
