Program HOC24;

var a: array[1..1000] of integer;

i,n,d: integer;

tbc: real;

t: longint;

begin

write('Nhap N: '); readln(n);

for i:=1 to n do 

begin

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

end;

t:=0;

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

writeln('Tong cac phan tu cua mang la: ',t);

d:=0; 

for i:=1 to n do if a[i]=5 then d:=d+1;

writeln('Co ',d,' phan tu co gia tri bang 5');

tbc:=0;

for i:=1 to n do if a[i] mod 2=1 then tbc:=tbc+a[i];

tbc:=tbc/n;

writeln('Trung binh cong cac phan tu le la: ',tbc:6:2);

write('Mang sau khi thay la: ');

for i:=1 to n do if a[i]=0 then a[i]:=10;

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

readln

end.

1,

\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)

\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)

\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)

\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)

\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)

\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)

4,

\(\frac{X+1}{5}=\frac{3}{7}\)

\(\Rightarrow\left(X-1\right).7=3.5\)

\(\Rightarrow7X-7=15\)

\(\Rightarrow7X=22\)

\(\Rightarrow X=\frac{22}{7}\)

\(\frac{X-3}{4}=\frac{1}{2}\)

\(\Rightarrow\left(X-3\right)2=1.4\)

\(\Rightarrow2X-6=4\)

\(\Rightarrow2X=10\)

\(\Rightarrow X=5\)

Tôi ko bt

Giúp mình đi

\(\dfrac{3}{28}\) ≤   \(\dfrac{x}{56}\) ≤ \(\dfrac{1}{4}\)

\(\dfrac{6}{56}\) ≤    \(\dfrac{x}{56}\) ≤ \(\dfrac{14}{56}\)

6 ≤        \(x\)    ≤ 14

Vì \(x\) nguyên nên \(x\) \(\in\) {6; 7; 8; 9; 10; 11; 12; 13; 14}

Vậy \(x\) \(\in\) {6; 7; 8; 9; 10; 11; 12; 13; 14}

cho cùng mẫu hả bạn

\(\frac{28}{15}\times\frac{1}{4}\times3+\left(\frac{8}{15}-\frac{79}{60}\right)\times\frac{24}{47}\)

=\(\frac{7}{15}\times3+\left(\frac{32}{60}-\frac{79}{60}\right)\times\frac{24}{47}\)

=\(\frac{7}{5}+\frac{47}{60}\times\frac{24}{47}\)

=\(\frac{7}{5}+\frac{2}{5}\)

=\(\frac{5}{5}\)

=\(1\)

\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{8}{15}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)=

=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+(\(\frac{32}{60}\)+\(\frac{-79}{60}\))x\(\frac{24}{47}\)

=\(\frac{28}{15}\)x\(\frac{1}{4}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)

=\(\frac{7}{15}\)x3+\(\frac{-47}{60}\)x\(\frac{24}{47}\)

=\(\frac{7}{5}\)+\(\frac{-2}{5}\)

=\(\frac{5}{5}\)=1