მათემატიკის წერთ რა ჯოჯოხეთი

program xoa_pt_trung_nhau;
uses crt;
var a:array[1..100] of real;
b:array[1..100] of boolean;
n,i,j:byte;
begin
clrscr;
write('nhap so luong phan tu cua mang: ');
readln(n);
for i:= 1 to n do
begin
write('nhap phan tu thu [',i,']= ');
readln(a[i]);
end;
for i:= 1 to n do
b[i]:=true;
for i:= 1 to n-1 do
for j:= i+1 to n do
if a[i]= a[j] then b[j]:=false;
for i:= 1 to n do
if b[i] =true then
write(a[i]:6:2);
readln
end.

program boi_cua_3;

uses crt;

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

n,i:byte;

s:integer;

begin

clrscr;

write('nhap so luong phan tu cua mang: ');

readln(n);

for i:= 1 to n do

begin

write('nhap phan tu thu [',i,']= ');

readln(a[i]);

end;

s:=0;

for i:= 1 to n do

if a[i] mod 3 = 0 then s:=s+a[i];

writeln('vay tong cac so la boi cua 3 la: ',s);

readln;

end.

program boi_cua_3;

uses crt;

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

n,i:byte;

min:integer;

begin

clrscr;

write('nhap so luong phan tu cua mang: ');

readln(n);

for i:= 1 to n do

begin

write('nhap phan tu thu [',i,']= ');

readln(a[i]);

end;

min:=a[1];

for i:= 1 to n do

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

writeln('gia tri nho nhat cua day la: ',min);

readln

end.

program baitap;

uses crt;

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

i,j,dem,dem2,n:byte;

S,T,T2:integer;

begin

clrscr;

write('nhap so phan tu cua mang: ');

readln(n);

for i:= 1 to n do

begin

write('nhap A[',i,']=');

readln(a[i]);

end;

cau a)===========================

S:=0

for j:= 1 to n do

if a[i] > 0 then

S:= S+a[i];

writeln('tong cac so duong trong day la: ',S);

cau b)=========================

T:=0;

dem:=0;

for i:= 1 to n do

if a[i] mod 2 = 0 then

begin

T:= T+a[i];

dem:=dem+1;

end;

writeln('Vay tong cac so chan co trong day la: ',T);

writeln('co',dem,'so chan co trong day ');

cau c)================================

T2:=0;

dem2:=0;

for i:= 1 to n do

if a[i] mod 3 = 0 then

begin

T2:= T2+a[i];

dem2:=dem2+1;

end;

writeln('Vay tong cac so chia het cho 3 co trong day la: ',T2);

writeln('co',dem2,'so chia het cho 3 co trong day ');

end.

program baitap;

uses crt;

var a,c,b,s:integer;

begin

clrscr;

write('nhap so thu nhat: ');

readln(a);

write('nhap so thu hai: ');

readln(b);

write('nhap so thu ba: ');

readln(c);

write(a:4,b:4,c:4);{xuat ra 3 so vua nhap}

writeln;

S:= a+b+c;

if S mod 5=0 then

writeln('tong ba so chia het cho 5')

else if S mod 3 = 0 then

writeln('tong ba so chia het cho 3');

readln;

end.

program fibonacci;

uses crt;

var a:array[0..1000] of integer;

i,n: integer;

begin

clrscr;

a[0] :=0;a[1]=1;

for i:= 2 to n do

a[i]=a[i-1]+a[i-2];

writeln('so hang thu ',n,' cua day la: ',a[n]);

end.

sửa chút để P max <=> x+25/x min nhé :))

a) \(Rđ=\dfrac{Uđm^2}{Pđm}=\dfrac{3^2}{3}=3\)

Rx=3Ω => R1 nt Rđ nt Rx

=> Rtd= R1+Rđ+Rx=2+3+3=8Ω

=> I=Iđ= U/Rtd=0.75(A)

=> P đèn = Rđ . I^2= 3.0.75^2=1.6875(W)

b) để đèn sáng bt <=> I=Iđm=Pđm/Uđm=3/3=1(A)

Rtd= 2+3+x=5+x

\(I=\dfrac{U}{Rtd}=\dfrac{6}{5+x}=1\) => x=1

=> Rx= 1Ω

c) ta có: Px=Rx.I^2=x.I^2

Rtd=5+x

\(I=\dfrac{6}{x+5}\) => \(Px=x.\left(\dfrac{6}{x+5}\right)^2=\dfrac{36x}{x^2+10x+25}=\dfrac{36}{x+10+\dfrac{25}{x}}\)

để Px max <=> x+ 25/x max

áp dụng bất đẳng thức cô si

\(x+\dfrac{25}{x}=5\) dấu '= ' sảy ra <=> x=25/x => x=5

=> Rx=5Ω => I= 0.6 (A) => Pmax=1.8(W)

((R2//R3)ntR1)//R4