Question 2 - End-of-Chapter Exercises - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)

Transcript

Question 2 (i) Find the values using suitable identities: (i) 17 × 21 Writing 17 × 21 = (19 – 2) × (19 + 2) So, we can write 𝟏𝟕 × 𝟐𝟏=(19−2) × (19+2) =〖𝟏𝟗〗^𝟐−𝟐^𝟐 =361−4 =𝟑𝟓𝟕 Using 𝑎^2−𝑏^2=(𝑎 −𝑏)(𝑎+𝑏) Putting 𝒂 = 𝟐𝟎, 𝒃 = 𝟑 Question 2 (ii) Find the values using suitable identities: (ii) 104×96 Writing 104 × 96 = (100 + 4) × (100 – 4) So, we can write 𝟏𝟎𝟒 × 𝟗𝟔=(100+4) × (100−4) =〖𝟏𝟎𝟎〗^𝟐−𝟒^𝟐 =10,000−16 =𝟗,𝟗𝟖𝟒 Using 𝑎^2−𝑏^2=(𝑎 −𝑏)(𝑎+𝑏) Putting 𝒂 = 𝟏𝟎𝟎, 𝒃 = 𝟒 Question 2 (iii) Find the values using suitable identities: (iii) 24×16 Writing 24 × 16 = (20 + 4) × (20 – 4) So, we can write 𝟐𝟒 × 𝟏𝟔=(20+4) × (20−4) =〖𝟐𝟎〗^𝟐−𝟒^𝟐 =400−16 =𝟑𝟖𝟒 Using 𝑎^2−𝑏^2=(𝑎 −𝑏)(𝑎+𝑏) Putting 𝒂 = 2𝟎, 𝒃 = 𝟒 Question 2 (iv) Find the values using suitable identities: (iv) 147^3 147^3=(𝟏𝟓𝟎−𝟑)^𝟑 =𝟏𝟓𝟎^𝟑−𝟑 × 𝟏𝟓𝟎^𝟐 × 𝟑+𝟑 × 𝟏𝟓𝟎 × 𝟑^𝟐−𝟑^𝟑 =3375000−3 × 22500 × 3+3 × 150 × 9−27 =3375000−202500+4050−27 =(3375000+4050)−(202500+27) =𝟑𝟏𝟕𝟔𝟓𝟐𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 150, 𝑏 = 3 Question 2 (v) Find the values using suitable identities: (v) 199^3 199^3=(𝟐𝟎𝟎−𝟏)^𝟑 =〖𝟐𝟎𝟎〗^𝟑−𝟑 × 𝟐𝟎𝟎^𝟐 × 𝟏+𝟑 × 𝟐𝟎𝟎 × 𝟏^𝟐−𝟏^𝟑 =8000000−3 × 40000 × 1+3 × 200 × 1−1 =8000000−120000+600−1 =(8000000+600)−(120000+1) =𝟕𝟖𝟖𝟎𝟓𝟗𝟗 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 200, 𝑏 = 3 Question 2 (vi) Find the values using suitable identities: (vi) 127^3 127^3=(𝟏𝟑𝟎−𝟑)^𝟑 =𝟏𝟑𝟎^𝟑−𝟑 × 𝟏𝟑𝟎^𝟐 × 𝟑+𝟑 × 𝟏𝟑𝟎 × 𝟑^𝟐−𝟑^𝟑 =2197000−3 × 16900 × 3+3 × 130 × 9−27 =2197000−152100+3510−27 =(2197000+3510)−(152100+27) =𝟐𝟎𝟒𝟖𝟑𝟖𝟑 Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 130, 𝑏 = 3 Question 2 (vii) Find the values using suitable identities: (vii) (−107)^3 Taking negative sign out (−𝟏𝟎𝟕)^𝟑=(−1 × 107)^3 =〖(−1)^3 × (107)〗^3 =〖−(107)〗^3 =〖−(𝟏𝟎𝟎+𝟕)〗^𝟑 =− [𝟏𝟎𝟎^𝟑+𝟑 × 𝟏𝟎𝟎^𝟐 × 𝟕+𝟑 × 𝟏𝟎𝟎 × 𝟕^𝟐+𝟕^𝟑] Using (𝑎+𝑏)^3=𝑎^3+3𝑎^2 𝑏+3𝑎𝑏^2+𝑏^3 Putting 𝑎 = 100, 𝑏 = 7 =− [1000000+3 ×10000 × 7+3 × 100 × 49+343] =− [1000000+210000 +14700 +343] =−𝟏𝟐𝟐𝟓𝟎𝟒𝟑 Question 2 (viii) Find the values using suitable identities: (viii) (−299)^3 Taking negative sign out (−𝟐𝟗𝟗)^𝟑=(−1 × 299)^3 =〖(−1)^3 × (299)〗^3 =〖−(299)〗^3 =〖−(𝟑𝟎𝟎−𝟏)〗^𝟑 =− [𝟑𝟎𝟎^𝟑−𝟑 × 𝟑𝟎𝟎^𝟐 × 𝟏+𝟑 × 𝟑𝟎𝟎 × 𝟏^𝟐−𝟏^𝟑] Using (𝑎−𝑏)^3=𝑎^3−3𝑎^2 𝑏+3𝑎𝑏^2−𝑏^3 Putting 𝑎 = 300, 𝑏 = 1 =− [27000000−3 × 90000 × 1+3 × 300 × 1−1] =− [27000000−270000 +900 −1] =− [(27000000+900)−(270000+1)] =− [27000900−270001] =−𝟐𝟔𝟕𝟑𝟎𝟖𝟗𝟗