\(\overrightarrow{AB}=\left(1-a;b;0\right);\overrightarrow{AC}=\left(1-a;0;c\right);\overrightarrow{HC}=\left(-2;-2;c-1\right);\overrightarrow{HB}=\left(-2;b-2;-1\right)\)

\(\left\{{}\begin{matrix}\overrightarrow{AB}.\overrightarrow{HC}=0\\\overrightarrow{AC}.\overrightarrow{HB}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-2\left(1-a\right)-2b=0\\-2\left(1-a\right)-c=0\end{matrix}\right.\) \(\Rightarrow c=2b=2\left(a-1\right)\) 

\(\left[\overrightarrow{AB};\overrightarrow{AC}\right]=\left(bc;c\left(a-1\right);b\left(a-1\right)\right)=\left(2\left(a-1\right)^2;2\left(a-1\right)^2;\left(a-1\right)^2\right)=\left(a-1\right)^2.\left(2;2;1\right)\)

A;B;C;H đồng phẳng 

\(\Rightarrow\left[\overrightarrow{AB};\overrightarrow{AC}\right].\overrightarrow{HB}=0\Rightarrow2.\left(-2\right)+2.\left(-2\right)+1.\left(c-1\right)=0\)

\(\Rightarrow c=9\Rightarrow b=\dfrac{9}{2}\Rightarrow a=\dfrac{11}{2}\)

Đáp án đúng : D

program bai1;
uses crt;
var i:integer;
s,s1:string;
begin
  clrscr;
  write('nhap S:');readln(s);
  while pos('C',s)<>0 do
  begin
    insert('LOP11A',s,pos('C',s));
    delete(s,pos('C',s),1);
  end;
  writeln('xau sau khi bien doi la: ',s);
  writeln('do dai cua xau tren la: ',length(s));
  write('nhap s1:');readln(s1);
  if s1[1]=s[1] then writeln('ki tu dau cua hai xau trung nhau')
  else writeln('ki tu dau cua hai xau khong trung nhau');
  readln;
end.

7,51x3=22,53

a=7

b=5

c=1

m=2

n=3

câu b: vì số này chia hết cho 2 và 5 nên b =0 còn a chắc bạn biết :D

Ko phai la chia het cha 2 va 5 mà là chia 5 và 9 du 3