101x 99 using algebraic identity
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Answered by
4
101×99
= (100+1)(100-1)
=(100×100)-(1×1)
=10000-1
=9999
here the identity used is (a+b) (a-b)= a^2- b^2
= (100+1)(100-1)
=(100×100)-(1×1)
=10000-1
=9999
here the identity used is (a+b) (a-b)= a^2- b^2
Answered by
5
We know that,
( a + b ) ( a - b ) = a² - b²
➡ so, 101 = 100 +1
➡ & 99 = 100 - 1
we have to find out,
101 * 99 = ( 100 + 1 ) * ( 100- 1 )
➡ 100² - 1²
➡ 10000 - 1
➡ 9999.
( a + b ) ( a - b ) = a² - b²
➡ so, 101 = 100 +1
➡ & 99 = 100 - 1
we have to find out,
101 * 99 = ( 100 + 1 ) * ( 100- 1 )
➡ 100² - 1²
➡ 10000 - 1
➡ 9999.
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