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

Last updated at March 22, 2023 by

Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 3

Ex 9.5, 5 - Chapter 9 Class 8 Algebraic Expressions and Identities - Part 4

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Ex 9.5, 5 Show that. (iii) (4/3 π‘šβˆ’3/4 𝑛)^2+2π‘šπ‘›=16/9 π‘š^2+9/16 𝑛^2 Solving LHS (4/3 π‘šβˆ’3/4 𝑛)^2+2π‘šπ‘› (π‘Žβˆ’π‘)^2=π‘Ž^2+𝑏^2βˆ’2π‘Žπ‘ Putting π‘Ž = 4/3 π‘š & 𝑏 = 3/4 𝑛 = (4/3 π‘š)^2+(3/4 𝑛)^2βˆ’2(4/3 π‘š)(3/4 𝑛)+2π‘šπ‘› = (4/3)^2Γ—π‘š^2+(3/4)^2×𝑛^2βˆ’2π‘šπ‘›+2π‘šπ‘› = 16/9 π‘š^2+9/16 𝑛^2 = RHS Since LHS = RHS Hence proved

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