Aspire Faculty ID #19208 · Topic: NIMCET 2026 · Just now
NIMCET 2026

The simplified Boolean value of the expression $(x+y'+z')(x+y'+z)(x+y+z')$ is:

Solution

Given expression is

$(x+y'+z')(x+y'+z)(x+y+z')$

Using the identity

$(x+A)(x+B)=x+AB$

First take

$(x+y'+z')(x+y'+z)$

Here,

$A=y'+z'$ and $B=y'+z$

So,

$(x+y'+z')(x+y'+z)=x+(y'+z')(y'+z)$

Now,

$(y'+z')(y'+z)=y'+z'z$

Since

$z'z=0$

So,

$(y'+z')(y'+z)=y'$

Now the expression becomes

$(x+y')(x+y+z')$

Again using the identity,

$(x+A)(x+B)=x+AB$

Here,

$A=y'$ and $B=y+z'$

So,

$(x+y')(x+y+z')=x+y'(y+z')$

$=x+y'y+y'z'$

Since

$y'y=0$

So,

$=x+y'z'$

Therefore, the simplified value is

$x+y'z'$

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