\(a,-\dfrac{10}{21}=\dfrac{-2}{7}.\dfrac{5}{3}=\dfrac{-2}{3}.\dfrac{5}{7}\\ b,\dfrac{2}{21}=\dfrac{1}{3}.\dfrac{2}{7}=\dfrac{1}{7}.\dfrac{2}{3}\\ c,\dfrac{8}{15}=\dfrac{2}{3}.\dfrac{4}{5}=\dfrac{4}{3}.\dfrac{2}{5}\)

a. \(\dfrac{10}{21}>1\)

\(\dfrac{9}{23}< 1\)

\(\Rightarrow\dfrac{10}{21}>\dfrac{9}{23}\)

b. \(\dfrac{32}{33}>\dfrac{31}{33}>\dfrac{31}{34}\Rightarrow\dfrac{32}{33}>\dfrac{31}{34}\)

c. \(\dfrac{44}{47}< \dfrac{45}{47}< \dfrac{45}{46}\Rightarrow\dfrac{44}{47}< \dfrac{45}{46}\)

\(d.\dfrac{70}{117}< \dfrac{70}{115}=\dfrac{14}{23}\Rightarrow\dfrac{70}{117}< \dfrac{14}{23}< \dfrac{15}{23}\)

\(\Rightarrow\dfrac{70}{117}< \dfrac{15}{23}\)

Mình không biết vì mình chỉ mới lên lớp 6 thôi ! Hì hì !

+Ta có: 50 = 2¹.5² => 50 có: (1 + 1).(2 + 1) = 6 (ước)

+Trong các số 10;11;21;31;51 có 2 số nguyên tố là 11 và 31

+Ta có: 450 = 2¹.3².5² => 450 có: (1 + 1).(2 + 1).(2 + 1) = 18 (ước)

+|8| + |-47| + |-69| = 8 + 47 + 69 = 124

ko biet thi thoi di zay cung noi

BCNN(50;8;20)=200

B(200)={0;200;400;600;800;1000;....}

Số nguyên dương có 3 chữ số là {200;400;600;800}

uses crt;

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

i,n,dem,dem1,dem2,t:integer;

s:real;

begin

clrscr;

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

for i:=1 to n do 

begin

repeat

write('A[',i,']='); readln(a[i]);

until a[i]>0;

end;

for i:=1 to n do 

  write(a[i]:4);

writeln;

dem:=0;

for i:=1 to n do

  if a[i]>10 then inc(dem);

writeln('So phan tu lon hon 10 la: ',dem);

dem1:=0;

dem2:=0;

for i:=1 to n do 

begin

if a[i] mod 2=0 then inc(dem1)

else inc(dem2);

end;

writeln('So luong so chan la: ',dem1);

writeln('So luong so le la: ',dem2);

t:=0;

s:=1;

for i:=1 to n do 

  begin

if (i mod 2=0) and (a[i] mod 2<>1) then t:=t+a[i];

if (i mod 2=1) and (a[i] mod 2=0) then s:=s*a[i];

end;

writeln('Tong cac so o vi tri chan co gia tri le la: ',t);

writeln('Tich cac so o vi tri le co gia tri chan la: ',s:4:2);

writeln('Cac so le la: ');

for i:=1 to n do 

  if a[i] mod 2<>0 then write(a[i]:4);

writeln;

writeln('Cac so chan va lon hon 10 la: ');

for i:=1 to n do 

  if (a[i] mod 2=0) and (a[i]>10) then write(a[i]:4);

readln;

end.