Var
A,B,X,Y:real;
Begin
Write('X = ');Read(X);
Write('Y = ');Read(Y);
Write('A = ');Read(A);
Write('B = ');Read(B);
if ((A<X)and(B<Y))or((A<Y)and(B<X)) then Write('Пройдёт')
else Write('Не пройдёт')
End.
Var
A,B,C:real;
Begin
Write('A = ');Read(A);
Write('B = ');Read(B);
if A>B then
Begin
C:=(A+B)/2;
B:=2*A*B;
A:=C;
End
else
Begin
C:=(A+B)/2;
A:=2*A*B;
B:=C;
End;
WriteLn('A = ',A);
WriteLn('B = ',B);
End.
Var X0,Y0,X1,Y1:byte;
Begin
Write('X0 = ');
Read(X0);
Write('Y0 = ');
Read(Y0);
Write('X1 = ');
Read(X1);
Write('Y1 = ');
Read(Y1);
if (((abs(X0-X1)=1)and(abs(Y0-Y1)=2))or((abs(X0-X1)=2)and(abs(Y0-Y1)=1))) then Write('Может')
else Write('Не может')
End.
Ответ:
3 будет
Объяснение:
Надо к конъюкции прибавить инверцию
Var a,b,c,d:integer;
begin
read(a,b);
c:=a+b;
d:=a*b;
if a < b then begin
a:=c;
b:=d;
end else begin
a:=d;
b:=c;
end;
write('a=',a,' b=',b);
end.
#include "stdio.h"
#include "math.h"
int main(){
double y, x, a, b;
scanf("%lf%lf%lf", &x, &a, &b);
y = cos(2 * x) + sin(x - 3);
printf("y = %lf", y);
if (cos(2 * a) + sin(x - a) > cos(2 * b) + sin(b - 3))
printf("\nmax - a");
else if (cos(2 * a) + sin(x - a) < cos(2 * b) + sin(b - 3))
printf("\nmax - b");
else
printf("\nin a = in b");
}
Если тебе нужен код, то вот он:
var
<span> x : real; </span>
<span>begin </span>
<span> write('x = ');readln(x); </span>
<span> if x > 0 </span>
<span> then </span>
<span> x := x - 50 </span>
<span> else </span>
<span> x := x + 100; </span>
<span> writeln('result : x = ',x:0:2); </span>
<span>end.</span>
