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प्रश्न
Expand `("a"-1/"a")^2`
बेरीज
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उत्तर
It is known that, (a + b)2 = a2 + 2ab + b2 and (a − b)2 = a2 − 2ab + b2.
`("a"-1/"a")^2`
= `("a")^2-2xx("a")xx(1/"a")+(1/"a")^2`
= `"a"^2-2+1/"a"^2`
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या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
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