F = !B & ( A + !(A&B) ) = !B & (A + !A + !B) = !B & (1+!B)= !B
{ т.к. !(A&B) = !A + !B }
F = !(A&B) + A&!B + A&B + B&C = !A + !B + A + B&C = !B + B&C
{ т.к. !(A&B) = !A + !B, A&!B + A&B = A&(!B+B) = A }
F = !(A&B + B&C) + C&!A = (A&B + B&C) & !(C&!A) =
= (A&B + B&C) & (!C+A) = B&(A+C)&(A+!C) = B & A
{ т.к. (A+C)&(A+!C) = AA + A!C + CA + C!C = A + A(!C+C)+0 = A + A = A }
Pascal
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 var a,b,c,s:real; begin repeat write ('Введите длину гипотенузы c='); readln(c); write ('Введите длину катета a='); readln(a); write ('Введите длину площади s='); readln(s); if (c>b) and (c>0) and (b>0) then begin a:=sqrt(sqr(c)-sqr(b)); s:=(1/2)*b*c; write end.
Const
n = 5;
var
arr: array [1..n] of integer;
i, max: integer;
begin
for i := 1 to n do
begin
write('A[', i, '] = ');
readln(arr[i]);
end;
max := 1;
for i := 1 to n do
if (max < arr[i]) and (arr[i] mod 5 = 0) then
max := arr[i];
if max = 1 then
writeln('No')
else
writeln(max);
end.
На горбушку) там много мастеров)