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

Advertisement

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

Advertisement

  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

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

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
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.