# Ex 9.5, 5 (iii) - Chapter 9 Class 8 Algebraic Expressions and Identities

Last updated at Sept. 8, 2021 by Teachoo

Last updated at Sept. 8, 2021 by Teachoo

Transcript

Ex 9.5, 5 Show that. (iii) (4/3 πβ3/4 π)^2+2ππ=16/9 π^2+9/16 π^2 Solving LHS (4/3 πβ3/4 π)^2+2ππ (πβπ)^2=π^2+π^2β2ππ Putting π = 4/3 π & π = 3/4 π = (4/3 π)^2+(3/4 π)^2β2(4/3 π)(3/4 π)+2ππ = (4/3)^2Γπ^2+(3/4)^2Γπ^2β2ππ+2ππ = 16/9 π^2+9/16 π^2 = RHS Since LHS = RHS Hence proved

Ex 9.5

Ex 9.5, 1 (i)
Important

Ex 9.5, 1 (ii)

Ex 9.5, 1 (iii)

Ex 9.5, 1 (iv)

Ex 9.5, 1 (v)

Ex 9.5, 1 (vi)

Ex 9.5, 1 (vii)

Ex 9.5, 1 (viii)

Ex 9.5, 1 (ix)

Ex 9.5, 1 (x)

Ex 9.5, 2 (i)

Ex 9.5, 2 (ii)

Ex 9.5, 2 (iii)

Ex 9.5, 2 (iv)

Ex 9.5, 2 (v)

Ex 9.5, 2 (vi)

Ex 9.5, 2 (vii)

Ex 9.5, 3 (i)

Ex 9.5, 3 (ii)

Ex 9.5, 3 (iii)

Ex 9.5, 3 (iv)

Ex 9.5, 3 (v)

Ex 9.5, 3 (vi)

Ex 9.5, 4 (i)

Ex 9.5, 4 (ii)

Ex 9.5, 4 (iii)

Ex 9.5, 4 (iv)

Ex 9.5, 4 (v)

Ex 9.5, 4 (vi)

Ex 9.5, 4 (vii)

Ex 9.5, 5 (i) Important

Ex 9.5, 5 (ii)

Ex 9.5, 5 (iii) You are here

Ex 9.5, 5 (iv)

Ex 9.5, 5 (v)

Ex 9.5, 6 (i) Important

Ex 9.5, 6 (ii)

Ex 9.5, 6 (iii)

Ex 9.5, 6 (iv)

Ex 9.5, 6 (v)

Ex 9.5, 6 (vi)

Ex 9.5, 6 (vii)

Ex 9.5, 6 (viii)

Ex 9.5, 6 (ix)

Ex 9.5, 7 (i)

Ex 9.5, 7 (ii)

Ex 9.5, 7 (iii)

Ex 9.5, 7 (iv)

Ex 9.5, 8 (i) Important

Ex 9.5, 8 (ii)

Ex 9.5, 8 (iii)

Ex 9.5, 8 (iv)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.