Ex 9.5

Chapter 9 Class 8 Algebraic Expressions and Identities
Serial order wise

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Ex 9.5, 4 Simplify. (v) (2.5πβ1.5π)^2 β(1.5πβ2.5π)^2 Solving (π.ππβπ.ππ)^π (πβπ)^2=π^2+π^2β2ππ Putting π = 2.5π & π = 1.5π = (2.5π)^2+(1.5π)^2β 2(2.5π)(1.5π) = (γ2.5γ^2Γπ^2 )+(γ1.5γ^2Γπ)^2 β (2Γ2.5Γ1.5)Γ(πΓπ) = (25/10)^2 π^2+(15/10)^2 π^2 β (2Γ25/10Γ15/10)ππ = 625/100 π^2+225/100 π^2β750/100 ππ Solving (π.ππβπ.ππ)^π (πβπ)^2=π^2+π^2β2ππ Putting π = 1.5π & π = 2.5π = (1.5π)^2+(2.5π)^2β2(1.5π)(2.5π) = (γ1.5γ^2Γπ^2 )+(γ2.5γ^2Γπ)^2β(2Γ1.5Γ2.5)Γ(πΓπ) = (15/10)^2 π^2+(25/10)^2 π^2β(2Γ15/10Γ25/10)ππ = 225/100 π^2+625/100 π^2β750/100 ππ β΄ (2.5πβ1.5π)^2 β(1.5πβ2.5π)^2 = 625/100 π^2+225/100 π^2β750/100 ππβ(225/100 π^2+625/100 π^2β750/100 ππ" " ) = 625/100 π^2+225/100 π^2β750/100 ππβ225/100 π^2β625/100 π^2+750/100 ππ = (625/100β225/100) π^2+(225/100β625/100) π^2+(β750/100 +750/100)ππ = 400/100 π^2β400/100 π^2 = ππ^πβππ^π