will go

A

a

uses crt;

const fi='dayso.inp',

var a:array[1..100]of integer;

i,n:integer;

f1:text;

begin

clrscr;

assign(f1,fi); reset(f1);

readln(f1,n);

for i:=1 to n do readln(f1,a[i]);

for i:=1 to n do write(a[i]:4);

close(f1);

readln;

end.

\(i=120^o-90^o=30^o\)

\(i=i'\Leftrightarrow i'=30^o\)

\(3n+1⋮11-n\)

\(=>3n+1⋮-\left(n-11\right)\)

\(=>3n-33+34⋮n-11\)

\(=>34⋮n-11\)

\(=>n-11\inƯ\left(34\right)\)

Nên ta có bảng sau :

Tự lập bảng nhé bạn :P

uses crt;

const fi='dst.txt'

var f1:text;

a,b:integer;

begin

clrscr;

assign(f1,fi); reset(f1);

readln(f1,a);

readln(f1,b);

writeln('Tong cua hai so la: ',a+b);

writeln('Gia tri tuyet doi hieu hai so la: ',abs(a-b));

readln;

end.

a) \(\Rightarrow\left(x-3\right)\left(x+4\right)=5.12\)

\(\Rightarrow x^2+x-72=0\)

\(\Rightarrow\left(x-8\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

b) \(\Rightarrow\left(x+3\right)^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-9\end{matrix}\right.\)

c) \(\Rightarrow2x^2=8\Rightarrow x^2=4\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

1:

uses crt;

const fi='songuyen.doc'

var i:integer;

f1:text;

begin

clrscr;

assign(f1,fi); rewrite(f1);

for i:=1 to 10 do write(f1,i:4);

close(f1);

readln;

end.

a) Đặt \(a=x^2+x\)

Đa thức trở thành: \(a^2-14a+24=\left(a^2-14a+49\right)-25=\left(a-7\right)^2-25=\left(a-7-5\right)\left(a-7+5\right)=\left(a-12\right)\left(a-2\right)\)

Thay a:

\(\left(a-12\right)\left(a-2\right)=\left(x^2+x-12\right)\left(x^2+x-2\right)\)

b) Đặt \(a=x^2+x\)

Đa thức trở thành:

\(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)-12=a^2+4a-12=\left(a^2+4x+4\right)-16=\left(a+2\right)^2-16=\left(a+2-4\right)\left(a+2+4\right)=\left(a-2\right)\left(a+6\right)\)

Thay a:

\(\left(a-2\right)\left(a+6\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)\)

Bài 1. x^2 \(\equiv\)8 (mod 0,1). (cmdd)

T tự: y^2 \(\equiv\)8 (mod 0,1)

=> x^2+y^2 \(\equiv\)8 (mod 0,1,2)

Mà 8z+6 \(\equiv\)8 (mod 6)

=> đpcm