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Above question modification.

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Boolean expression$:(x+y)(x+\bar{y})+\overline{{(x\bar{y}+\bar{x})}}$

$(A) x$               $(B)y$           $(C)xy$               $(D)x+y$      

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$f=(x+y)(x+\bar y)+\overline{(x\bar y+\bar x)}$

$f=(x+x\bar y+xy+y\bar y)+\overline{x\bar y}.\bar{ \bar x}$

$f=(x(1+\bar y+y))+(\bar x+y).x$

$f=x+xy$

$f=x(1+y)=x$

Option (A) is correct.
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1 votes
$(x+y)(x+\overline{y}) + \overline{x\overline{y} + \overline{x} }$

$xy + x\overline{y} + \overline{x\overline{y}}$ $  (\overline{\overline{x}}) $

$x + (\overline{x} + y)x$

$x + yx$

$x$
1 votes
1 votes

Answer : A

 

(x+y)(x+y')+(xy'+x')'

(x+y)(x+y')+((x'+y)(x))   //  (x'+y)(x) = xy

x+xy'+xy+xy     // x+xy'=x

x + xy    // x+xy = x

x

 

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