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

Transcript

Ex 9.5, 8 Using (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ , find (i) 103 Γ— 104 103 Γ— 104 = (100+3) Γ—(100+4) (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 100 , π‘Ž = 3 & 𝑏 = 4 (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 100 , π‘Ž = 3 & 𝑏 = 4 Ex 9.5, 8 Using (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ , find (ii) 5.1 Γ— 5.2 5.1 Γ— 5.2 = (5+0.1)Γ—(5+0.2) (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 5 , π‘Ž = 0.1 & 𝑏 = 0.2 = (5)^2+(0.1+0.2)(5)+(0.1)(0.2) = 25+(0.3)5+(0.1Γ—0.2) = 25+3/10Γ—5+(1/10Γ—2/10) = 25+(15/10)+2/100 = 25+1.5+0.02 = πŸπŸ”.πŸ“πŸ Ex 9.5, 8 Using (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ , find (iii) 103 Γ— 98 103 Γ—98 = (100+3)Γ—(100βˆ’2) = (100+3)Γ—[100+(βˆ’2)] (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 100 , π‘Ž = 3 & 𝑏 = βˆ’2 = (100)^2+(3+2)(100)+(3)(βˆ’2) = 10000+100βˆ’6 = 10000+94 = πŸπŸŽπŸŽπŸ—πŸ’ Ex 9.5, 8 Using (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ , find (iv) 9.7 Γ— 9.8 9.7 Γ—9.8 = (10βˆ’0.3)Γ—(10βˆ’0.2) = (10+(βˆ’0.3))Γ—[10+(βˆ’0.2)] (π‘₯+π‘Ž)(π‘₯+𝑏)=π‘₯^2+(π‘Ž+𝑏)π‘₯+π‘Žπ‘ Putting π‘₯ = 10 , π‘Ž = βˆ’0.3 & 𝑏 = βˆ’0.2 = (10)^2+(βˆ’0.3βˆ’0.2)(10)+(βˆ’0.3)(βˆ’0.2) = 100+(βˆ’0.5Γ—10)+(0.3Γ—0.2) = 100βˆ’(5/10Γ—10)+(3/10Γ—2/10) = 100βˆ’5+6/100 = 95+0.06 = πŸ—πŸ“.πŸŽπŸ”

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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.