Question

USING IDENTITIES, EVALUTE​

Answer 1

Step-by-step explanation:

we have to use algebraic identities to solve these questions

$( {x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\ ( {x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ \\ {x}^{2} - {y}^{2} = (x - y)(x + y)$

$1) \: {(71)}^{2} = ( {70 + 1)}^{2} \\ \\ = {(70)}^{2} + 2(70)(1) + {1}^{2} \\ \\ = 4900 + 140 + 1 \\ \\ {(71)}^{2} = 5041$

$2)( {99)}^{2} = {(100 - 1)}^{2} \\ \\ = ( {100)}^{2} - 2(100)(1) + ( {1)}^{2} \\ \\ = 10000 - 200 + 1 \\ \\ {( 99) }^{2} = 9801$

$3)( {102)}^{2} = {(100 + 2)}^{2} \\ \\ = ( {100)}^{2} + 2(100)(2) + ( {2)}^{2} \\ \\ = 10000 + 400 + 4 \\ \\ ( {102)}^{2} = 10404$

$4) {(998)}^{2} = ( {1000 - 2)}^{2} \\ \\ = ( {1000)}^{2} - 2(1000)(2) + ( {2)}^{2} \\ \\ = 1000000 - 4000 + 4 \\ \\ ( {998)}^{2} = 996004$

$5)( {5.2)}^{2} = ( {5 + 0.2)}^{2} \\ \\ = ( {5)}^{2} + 2(5)(0.2) + ( {0.2)}^{2 } \\ \\ = 25 + 2 + 0.04 \\ \\ ( {5.2)}^{2} = 27.04$

$6)297 \times 303 \\ \\ = (300 - 3)(300 + 3) \\ \\ = ( {300)}^{2} - ( {3)}^{2} \\ \\ = 90000 - 9 \\ \\ 297 \times 303= 89991$

$7)78 \times 82 = (80 - 2)(80 + 2) \\ \\ = ( {80)}^{2} - ( {2)}^{2} \\ \\ = 6400 - 4 \\ \\ 78 \times 82 = 6396$

$8)( {8.9)}^{2} = ( {9 - 0.1)}^{2} \\ \\ = ( {9)}^{2} - 2(0.1)(9) + ( {0.1)}^{2} \\ \\ = 81 - 1.8 + 0.01 \\ \\ ( {8.9)}^{2} = 79.21$

$9)10.5 \times 9.5 \\ \\ = (10 +0 .5)(10 - 0.5) \\ \\ = ( {10)}^{2} - ( {0.5)}^{2} \\ \\ = 100 - 0.25 \\ \\ = 99.75$

Hope it helps you.