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प्रश्न
Find the square of `"x"+1/"x"-1`
बेरीज
उत्तर
`("x"+1/"x"-1)^2=("x")^2+(1/"x")^2+(-1)^2+2xx"x"xx1/ "x"+2xx1/"x"xx(-1)+2(-1)xx"x"`
`="x"^2+1/"x"^2+1+2-2/"x"-2"x"`
`="x"^2+1/"x"^2+3-2/"x"-2"x"`
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