program amduong;

var n:int64;

begin

readln(n);

if n<0 then write('am') else if n>0 then write('duong') else write('khong phai so am cung khong phai so duong');

readln;

end.

program diem;

var st,s:ansistring;i,j:longint;n:real;

begin

readln(st);

for i:=1 to length(st) do if st[i] in ['0'..'9'] then

begin

for j:=i to length(st) do s:=s+st[j];

val(s,n);

if n>5 then write('dau') else write('rot');

break;

end;

readln;

end.

1:

uses crt;

var a:integer;

begin

clrscr;

write('nhap a='); readln(a);

if a>0 then writeln(a,' la so duong');

if a<0 then writeln(a,' la so am');

if a=0 then writeln(a,' khong la so am cung khong la so duong');

readln;

end.

2:

uses crt;

var st,s:string;

n:real;

i,j,d,x:integer;

begin

clrscr;

write('nhap ten:');readln(st);

s:='';

d:=length(st);

for i:=1 to d do

if st[i] in ['0'..'9'] then

begin

for j:=i to d do

s:=s+st[j];

val(s,n,x);

if n>=5 then writeln('dau')

else writeln('khong dau');

break;

end;

readln;

end.

b) Ta có: \(\left(-\dfrac{1}{2}\right)^2\cdot2\dfrac{6}{7}-\dfrac{14}{15}:2\dfrac{1}{3}+\left(-1.21\right)^0\)

\(=\dfrac{1}{4}\cdot\dfrac{20}{7}-\dfrac{14}{15}:\dfrac{7}{3}+1\)

\(=\dfrac{5}{7}-\dfrac{14}{15}\cdot\dfrac{3}{7}+1\)

\(=\dfrac{5}{7}-\dfrac{2}{5}+1\)

\(=\dfrac{25-14-35}{35}=\dfrac{-24}{35}\)

Sửa đề: Kiểm tra N là số chẵn hay số lẻ

uses crt;

var n:longint;

begin

clrscr;

repeat

write('Nhap n='); readln(n);

until n>0;

if n mod 2=0 then writeln(n,' la so chan')

else writeln(n,' la so le');

readln;

end.

bạn ghi rõ nội dung, yêu cầu đề của bài đó ra cho mình nha

b: 

#include <bits/stdc++.h>

using namespace std;

long long a,b,t;

int main()

{

cin>>a>>b;

t=a+b;

cout<<t;

return 0;

}

:)))

chịu .-. 

;-;

ai bt,ik mà hỏi ngta

b: 

#include <bits/stdc++.h>

using namespace std;

long long a,b,t;

int main()

{

cin>>a>>b;

t=a+b;

cout<<t;

return 0;

}