MATHS FLASH CARDS
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| 🇬🇧 | 🇬🇧 |
| (2x – 1/x)² | Solution: (i) (2x – 1/x)² [Use identity: (a – b)² = a² + b² – 2ab ] = (2x – 1/x)² = (2x)² + (1/x)² – 2 (2x)(1/x) = 4x² + 1/x² – 4 |
| (2x + y) (2x – y) | [Use identity: (a – b)(a + b) = a² – b² ] (2x + y) (2x – y) = (2x )² – (y)² = 4x² – y² |
| (a + 2b + c)² | (a + 2b + c)² = a² + (2b) ² + c² + 2a(2b) + 2ac + 2(2b)c = a² + 4b² + c² + 4ab + 2ac + 4bc |
| (2+x−2y)² | (2+x−2y)² = 2² + x² + (−2y)² + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y) = 4 + x² + 4y² + 4 x − 4xy − 8y |
| (2x−y+z) 2 | (2x−y+z) 2 = (2x) 2 + (−y) 2 + (z) 2 + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z) = 4x2 + y2 + z2 − 4xy−2yz+4xz |
| (2x−y+z) ² | (2x−y+z) ² = (2x) ² + (−y) ² + (z)² + 2(2x)(−y) + 2(−y)(z) + 2(2x)(z) = 4x² + y² + z²− 4xy−2yz+4xz |
| (−2x+3y+2z) ² | = (−2x) ² + (3y) ² + ( 2z) ²+ 2(−2x)(3y)+2(3y)(2z)+2(−2x)(2z) = 4x² + 9y² + 4z² −12xy+12yz−8xz |
| Factorize the given expression: 27x³ - 63x²+ 49x - 343/27 | Compare the given expression to the standard cubic algebraic identity to factorise (a - b)3 = a3 - 3a2b + 3ab2 - b3 Compare 27x3 - 63x2 + 49x - 34327 to a3 - 3a2b + 3ab2 - b3 We can deduce that 27x3 = a3 or (3x)3 = a3 ∴ a = 3x And 34327 = b3 or [73]3 = b3 ∴ b = 73 So, 27x3 - 63x2 + 49x - 34327 |
| (i) (1/x + y/3) | [Using identities: (a + b)³ = a³ + b³ + 3ab(a + b) and (a – b)³ = a³ – b³ – 3ab(a – b) ] |
| (ii) (3/x – 2/x2) | [Using identities: (a + b)3 = a3 + b3 + 3ab(a + b) and (a – b)3 = a3 – b3 – 3ab(a – b) ] |
| (ii) (4 – 1/3x) | [Using identities: (a + b)3 = a3 + b3 + 3ab(a + b) and (a – b)3 = a3 – b3 – 3ab(a – b) ] |
| (ii) (x/2 + y/3) ³ | (x/2 + y/3) ³ [Using identities: (a + b)³ = a³ + b³ + 3ab(a + b) and (a – b)³ = a³ – b³ – 3ab(a – b) Here a = (x/2 ) and b = ( y/3) x³/8 + y³ / 27 + 1/4 (3x²y + 2xy²) |
| Factorize the given expression: 27x³ - 63x²+ 49x - 343/27 | Compare the given expression to the standard cubic algebraic identity to factorise (a - b)³ = a³ - 3a²b + 3ab² - b³ Compare 27x³ - 63x² + 49x - 343/27 to a³ - 3a²b + 3ab² - b³ We can deduce that 27x³ = a³ or (3x)³ = a³ ∴ a = 3x And 343/27 = b³ or [73]³ = b³ ∴ b = 7/3 So, 27x³ - 63x²+ 49x - 343/27 (3x)³ - 3(3x)²(7/3) + 3(3x)[7/3]² - [7/3]³ = [3x−7/3]³ = [3x−7/3] [3x−7/3] [3x−7/3] |
| Using suitable identity evaluate the following: 98³ | (98)³ = (100 - 2)³ = (100 - 2)³is of the form (a - b)³ (a - b)³ = a³ - 3a²b + 3ab² - b³ So, (100 - 2)³ = 100³ - (3 × 100² × 2) + (3 × 100 × 2²) - 2³ = 1,000,000 - 60,000 + 1200 - 8 = 9,41,192 |