Slide30.JPG

Slide31.JPG
Slide32.JPG

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

Transcript

Example 13 Using Identity (III), find (i) (3/2 π‘š+2/3 𝑛)(3/2 π‘šβˆ’2/3 𝑛) (3/2 π‘š+2/3 𝑛)(3/2 π‘šβˆ’2/3 𝑛) (π‘Ž+𝑏)(π‘Žβˆ’π‘)=π‘Ž^2βˆ’π‘^2 Putting π‘Ž = 3/2 π‘š & 𝑏 = 2/3 𝑛 = (3/2 π‘š)^2βˆ’(2/3 𝑛)^2 = (3/2)^2Γ—π‘š^2βˆ’(2/3)^2×𝑛^2 = πŸ—/πŸ’ π’Ž^πŸβˆ’πŸ’/πŸ— 𝒏^𝟐 Example 13 Using Identity (III), find (ii) γ€–983γ€—^2βˆ’ γ€–17γ€—^2 γ€–983γ€—^2βˆ’ γ€–17γ€—^2 π‘Ž^2βˆ’π‘^2=(π‘Ž+𝑏)(π‘Žβˆ’π‘) Putting π‘Ž = 983 & 𝑏 = 17 = (983+17)Γ—(983βˆ’17) = 1000Γ—966 = 9,66,000 Example 13 Using Identity (III), find (iii) 194 Γ— 206 194 Γ— 206 = (200βˆ’6)Γ—(200+6) 194 Γ— 206 = (200βˆ’6)Γ—(200+6) = (200)^2βˆ’(6)^2 = 40000βˆ’36 = 39964

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.