Menu Close

Simplify-1-A-B-A-B-2-A-B-A-B-




Question Number 168858 by Shrinava last updated on 19/Apr/22
Simplify:  1. (A + B)(A + B^(−) )  2. (A^(−)  + B)(A^(−)  + B^(−) )
$$\mathrm{Simplify}: \\ $$$$\mathrm{1}.\:\left(\mathrm{A}\:+\:\mathrm{B}\right)\left(\mathrm{A}\:+\:\overline {\mathrm{B}}\right) \\ $$$$\mathrm{2}.\:\left(\overline {\mathrm{A}}\:+\:\mathrm{B}\right)\left(\overline {\mathrm{A}}\:+\:\overline {\mathrm{B}}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 19/Apr/22
1. (A + B)(A + B^(−) )       =AA+AB^(−) +BA+BB^(−)        =A+AB^(−) +AB+0       =A+A(B^(−) +B)       =A+A∙1       =A+A       =A  2. (A^(−)  + B)(A^(−)  + B^(−) )       =A^(−) A^(−) +A^(−) B^(−) +BA^(−) +BB^(−)        =A^(−) +A^(−) (B^(−) +B)+0       =A^(−) +A^(−) (1)       =A^(−) +A^(−)        =A^(−)
$$\mathrm{1}.\:\left(\mathrm{A}\:+\:\mathrm{B}\right)\left(\mathrm{A}\:+\:\overline {\mathrm{B}}\right) \\ $$$$\:\:\:\:\:=\mathrm{AA}+\mathrm{A}\overline {\mathrm{B}}+\mathrm{BA}+\mathrm{B}\overline {\mathrm{B}} \\ $$$$\:\:\:\:\:=\mathrm{A}+\mathrm{A}\overline {\mathrm{B}}+\mathrm{AB}+\mathrm{0} \\ $$$$\:\:\:\:\:=\mathrm{A}+\mathrm{A}\left(\overline {\mathrm{B}}+\mathrm{B}\right) \\ $$$$\:\:\:\:\:=\mathrm{A}+\mathrm{A}\centerdot\mathrm{1} \\ $$$$\:\:\:\:\:=\mathrm{A}+\mathrm{A} \\ $$$$\:\:\:\:\:=\mathrm{A} \\ $$$$\mathrm{2}.\:\left(\overline {\mathrm{A}}\:+\:\mathrm{B}\right)\left(\overline {\mathrm{A}}\:+\:\overline {\mathrm{B}}\right) \\ $$$$\:\:\:\:\:=\overline {\mathrm{A}}\overline {\mathrm{A}}+\overline {\mathrm{A}}\overline {\mathrm{B}}+\mathrm{B}\overline {\mathrm{A}}+\mathrm{B}\overline {\mathrm{B}} \\ $$$$\:\:\:\:\:=\overline {\mathrm{A}}+\overline {\mathrm{A}}\left(\overline {\mathrm{B}}+\mathrm{B}\right)+\mathrm{0} \\ $$$$\:\:\:\:\:=\overline {\mathrm{A}}+\overline {\mathrm{A}}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:=\overline {\mathrm{A}}+\overline {\mathrm{A}} \\ $$$$\:\:\:\:\:=\overline {\mathrm{A}} \\ $$
Commented by Shrinava last updated on 19/Apr/22
perfect thankyou sir
$$\mathrm{perfect}\:\mathrm{thankyou}\:\boldsymbol{\mathrm{sir}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *