Question

prove that by using situable identities....100-100/100-100=2 where 100-100 is not equal to 0​

Answer 1

Answer:

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Answer 2

$\mathfrak{dear\;user }$

$\mathfrak{question-}$prove 100-100/100-100=2 ​

$\mathfrak{here\:is\:the\:solution\:for \:the question }$

$a^{2}-b^{2} =(a+b)(a-b)$

$= (100-100)\div (100-100)=2$

$L.H.S$

=$(10^{2} -10^{2})\div10(10-10)$

=$(10+10)(10-10)\div10(10-10)$ $cancel$$(10-10)$$from \:numerator \:and \:denominatior.$

=$10+10\div10$

=$20 \div 10$

$=2$

$R.H.S=2$

$\mathcal{MY\:EXPECTATION\: FOR \:THIS \: ANSWER \:IS }10\:thanks\:and \:brainlist$

$\mathcal{BY \:ROSHAN\: A \:USER \: OF \: BRAINLY}$

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