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Ex 9.5, 5 (i) - Chapter 9 Class 8 Algebraic Expressions and Identities
Last updated at March 31, 2023 by Teachoo
Ex 9.5
Ex 9.5, 1 (i)
Ex 9.5, 1 (ii) Important
Ex 9.5, 1 (iii)
Ex 9.5, 1 (iv)
Ex 9.5, 1 (v) Important
Ex 9.5, 1 (vi) Important
Ex 9.5, 1 (vii)
Ex 9.5, 1 (viii)
Ex 9.5, 1 (ix) Important
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) Important
Ex 9.5, 2 (v)
Ex 9.5, 2 (vi) Important
Ex 9.5, 2 (vii) Important
Ex 9.5, 3 (i)
Ex 9.5, 3 (ii) Important
Ex 9.5, 3 (iii)
Ex 9.5, 3 (iv) Important
Ex 9.5, 3 (v) Important
Ex 9.5, 3 (vi)
Ex 9.5, 4 (i)
Ex 9.5, 4 (ii)
Ex 9.5, 4 (iii) Important
Ex 9.5, 4 (iv)
Ex 9.5, 4 (v) Important
Ex 9.5, 4 (vi)
Ex 9.5, 4 (vii) Important
Ex 9.5, 5 (i) You are here
Ex 9.5, 5 (ii)
Ex 9.5, 5 (iii) Important
Ex 9.5, 5 (iv)
Ex 9.5, 5 (v) Important
Ex 9.5, 6 (i)
Ex 9.5, 6 (ii) Important
Ex 9.5, 6 (iii)
Ex 9.5, 6 (iv)
Ex 9.5, 6 (v) Important
Ex 9.5, 6 (vi)
Ex 9.5, 6 (vii) Important
Ex 9.5, 6 (viii)
Ex 9.5, 6 (ix) Important
Ex 9.5, 7 (i)
Ex 9.5, 7 (ii) Important
Ex 9.5, 7 (iii)
Ex 9.5, 7 (iv) Important
Ex 9.5, 8 (i)
Ex 9.5, 8 (ii)
Ex 9.5, 8 (iii) Important
Ex 9.5, 8 (iv) Important
Transcript
Ex 9.5, 5 Show that. (i) (3π₯+7)^2β84π₯=(3π₯β7)^2 Solving LHS (3π₯+7)^2β84π₯ = (3π₯)^2+(7)^2+ 2(3π₯)(7)β84π₯ = 9π₯^2+49+42π₯β84π₯ = 9π₯^2+49β42π₯ (π+π)^2=π^2+π^2β2ππ Putting π = 3π₯ & π = 7 = (3π₯)^2+(7)^2+ 2(3π₯)(7)β84π₯ = 9π₯^2+49+42π₯β84π₯ = 9π₯^2+49β42π₯ Solving RHS (3π₯β7)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 3π₯ & π = 7 = (3π₯)^2+(7)^2β2(3π₯)(7) = (3^2Γπ₯^2 )+49β(2Γ3Γ7)π₯ = 9π₯^2+49β42π₯ Thus LHS = RHS Hence proved
