Advertisements
Advertisements
प्रश्न
Verify the following:
`((3p)/7 + 7/(6p))^2 - (3/7p + 7/(6p))^2 = 2`
Advertisements
उत्तर
Taking L.H.S. = `((3p)/7 + 7/(6p))^2 - (3/7p + 7/(6p))^2 = 2`
= `[((3p)/7 + 7/(6p)) + ((3p)/7 - 7/(6p))][((3p)/7 + 7/(6p)) - ((3p)/7 - 7/(p))]` ...[Using the identity, a2 – b2 = (a + b)(a – b)]
= `((3p)/7 + 7/(6p) + (3p)/7 - 7/(6p))((3p)/7 + 7/(6p) - (3p)/7 + 7/(6p))`
= `(6p)/7 xx 14/(6p)`
= 2
= R.H.S.
Hence verified.
APPEARS IN
संबंधित प्रश्न
Factorise : 16p4 – 1
Using the identity (a + b)(a – b) = a2 – b2, find the following product
(6x + 7y)(6x – 7y)
Factorise the following algebraic expression by using the identity a2 – b2 = (a + b)(a – b)
25a2 – 49b2
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
3a2b3 – 27a4b
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
49x2 – 36y2
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
`x^2/25 - 625`
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
`(x^3y)/9 - (xy^3)/16`
Factorise the following using the identity a2 – b2 = (a + b)(a – b).
p5 – 16p
Factorise the expression and divide them as directed:
(3x4 – 1875) ÷ (3x2 – 75)
Verify the following:
(p – q)(p2 + pq + q2) = p3 – q3
