346,25 x 12,9 =4466,625

var i,n,s:integer;

begin

readln(n); s:=0;

for i:=1 to 30 do

s:=s+i;

wirteln(s);

end.

1) var i,n:ineteger; s:real;

begin

readln(n); s:=0;

for i:=1 to n do

s:=s+2*i*(i+2);

writeln(s);

readln;

end.

Type mang=array [1..20] of integer;

var a:mang; i,n:integer;

begin

readln(n);

for i:=1 to n do

begin

readln(a[i]);

end;

for i:=1 to n do

if a[i] mod 2 = 0 then writeln(a[i])

else writeln(' khong in');

end.

2) type mang=array [1..20] of integer;

var a:mang; n,i,tong,tich:integer;

begin

readln(n); tong:=0; tich:=1

for i:=1 to n do

begin

readln(a[i]);

end;

for i:=1 to n do tong:=tong+a[i];

for i:=1 to n do tich:=tich*a[i];

end;

writeln(tong); writeln(tich);

readln;

end.

1) Type mang=array [1..20] of integer;

var a:mang; i,n,max,min:integer;

begin

readln(n);

for i:=1 to n dO

begin

readln(a[i]);

end;

max:=a[1]; min:=a[1];

for i:=2 to n do

begin

if max < a[i] then max:=a[i];

if min > a[i] then min:=a[i];

end;

writeln(min); writeln(max);

readln;

end.

a) Áp dụng tính chất của dảy tỉ số bằng nhau ta có:

\(\frac{1+2y}{18}=\frac{1+4y}{24}=\frac{1+6y}{6x}=\frac{1+2y+1+6y}{18+6x}=\frac{2+8y}{18+6x}\)

\(=\frac{2.\left(1+4y\right)}{2.\left(9+3x\right)}=\frac{1+4y}{9+3x}\)

suy ra: \(\frac{1+4y}{24}=\frac{1+4y}{9+3x}\Rightarrow9+3x=24\Rightarrow3x=15\Rightarrow x=5\)

b)Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x-1}{x+2}=\frac{x-2}{x+3}=\frac{\left(x-1\right)-\left(x-2\right)}{\left(x+2\right)-\left(x+3\right)}=\frac{x-1-x+2}{x+2-x-3}=\frac{1}{-1}=-1\)

Suy ra:

\(\frac{x-1}{x+2}=-1\Rightarrow x-1=-1.\left(x+2\right)\Rightarrow x-1=-x-2\)

\(\Rightarrow x+x=-2+1\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)

(8x - 1)²ⁿ⁺¹ = 5²ⁿ⁺¹

8x - 1 = 5

8x - 1 = 5

8x = 6

x = \(\frac{3}{4}\)

Tính & in tổng phần tử tại vị trí lẻ:

Type mang=array [1..20] of integer;

var a:mang; i,n,s:integer;

begin

s:=0;

readln(n);

for i:=1 to n do

begin

readln(a[i]);

end;

writeln('tong la');

for i:=1 to n do

if i mod 2 <>0 then s:=s+a[i];

writeln(s);

readln;

end.