1)
(X+Z+Y)(X+!Y+Z)(X+!Y+!Z) -> X(X+!Y+Z)(X+!Y+!Z)+Z(X+!Y+Z)(X+!Y+!Z)+Y(X+!Y+Z)(X+!Y+!Z)
X(X+!Y+Z)(X+!Y+!Z) -> XX(X+!Y+!Z)+!YX(X+!Y+!Z)+ZX(X+!Y+!Z)
XX(X+!Y+!Z) -> XX+!YX+!ZX
XX -> X
XX+!YX+!ZX -> X
!YX(X+!Y+!Z) -> X!YX+!Y!YX+!Z!YX
X!YX -> X!Y
!Y!YX -> !YX
X!YX+!Y!YX+!Z!YX -> X!Y
ZX(X+!Y+!Z) -> XZX+!YZX+!ZZX
XZX -> XZ
!ZZX -> 0
XZX+!YZX+!ZZX -> XZ
XX(X+!Y+!Z)+!YX(X+!Y+!Z)+ZX(X+!Y+!Z) -> X
Z(X+!Y+Z)(X+!Y+!Z) -> XZ(X+!Y+!Z)+!YZ(X+!Y+!Z)+ZZ(X+!Y+!Z)
XZ(X+!Y+!Z) -> XXZ+!YXZ+!ZXZ
XXZ -> XZ
!ZXZ -> 0
XXZ+!YXZ+!ZXZ -> XZ
!YZ(X+!Y+!Z) -> X!YZ+!Y!YZ+!Z!YZ
!Y!YZ -> !YZ
!Z!YZ -> 0
X!YZ+!Y!YZ+!Z!YZ -> !YZ
ZZ(X+!Y+!Z) -> XZ+!YZ+!ZZ
!ZZ -> 0
XZ+!YZ+!ZZ -> XZ+!YZ
XZ(X+!Y+!Z)+!YZ(X+!Y+!Z)+ZZ(X+!Y+!Z) -> XZ+!YZ
Y(X+!Y+Z)(X+!Y+!Z) -> XY(X+!Y+!Z)+!YY(X+!Y+!Z)+ZY(X+!Y+!Z)
XY(X+!Y+!Z) -> XXY+!YXY+!ZXY
XXY -> XY
!YXY -> 0
XXY+!YXY+!ZXY -> XY
!YY(X+!Y+!Z) -> 0
ZY(X+!Y+!Z) -> XZY+!YZY+!ZZY
!YZY -> 0
!ZZY -> 0
XZY+!YZY+!ZZY -> XZY
XY(X+!Y+!Z)+!YY(X+!Y+!Z)+ZY(X+!Y+!Z) -> XY
X(X+!Y+Z)(X+!Y+!Z)+Z(X+!Y+Z)(X+!Y+!Z)+Y(X+!Y+Z)(X+!Y+!Z) -> X+!YZ
2)
(A+!B)+A+FE+B -> 1
3)
!(X+T) -> !X!T
TYZ(!(X+T)) -> 0
X+TYZ(!(X+T)) -> X
0.01010001111010111000010100011110101110000101000<span>111 в двоичной с/c
</span><span><span>0.243656050753412172702436560507534121727024365605078 в восьмеричной с/с</span><span>
</span></span>
program slova;
uses crt;
var s, s1: string;
i, k: integer;
begin
clrscr;
writeln ('Vvedite text: ');
readln (s1);
write ('Slova soderjaschie ne bolee 3-h bukv: ');
for i:=1 to length(s) do
begin
if s[i]<>' ' then
begin
s1:=s1+s[i];
k:=k+1;
end
else
if k<=3 then
begin
write (', ',s1);
k:=0;
delete (s1,1,3);
end
else
begin
k:=0;
delete (s1,1,15);
end;
end;
readln;
end.
Вроде так, но у меня щас нету компилятора, так что проверить не могу. Если что-то будет ругаться, или вопросы какие - пиши в личку.
А) if (x >= m) and (x <= n) then written('True') else Writeln('False');
b) if (c*c = a*a + b*b) then written('True') else writeln('False');
в)if (x > 0) and (y > 0) then written('True') else writeln('False');
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Кнопка 'спасибо' чуть ниже