Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 9.5, 5 (v) - Chapter 9 Class 8 Algebraic Expressions and Identities
Last updated at March 22, 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)
Ex 9.5, 5 (ii)
Ex 9.5, 5 (iii) Important
Ex 9.5, 5 (iv)
Ex 9.5, 5 (v) Important You are here
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. (v) (πβπ) (π+π)+(πβπ)(π+π)+(πβπ)(π+π)=0 Solving LHS (πβπ) (π+π)+(πβπ)(π+π)+(πβπ)(π+π) Using (π₯βπ¦)(π₯+π¦)=π₯^2+π¦^2 = (π^2βπ^2 )+(π^2βπ^2 )+(π^2βπ^2 ) = π^2βπ^2+π^2βπ^2+π^2βπ^2 = (π^2βπ^2 )+(βπ^2+π^2 )+(βπ^2+π^2 ) = 0+0+0 = 0 = R.H.S Since LHS = RHS Hence proved
