Cho N bằng 1 phần 4 + 1 phần 28 + 1 phần 77 + 1 phần 54 tính N
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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,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
\(\dfrac{3}{28}\) ≤ \(\dfrac{x}{56}\) ≤ \(\dfrac{1}{4}\)
\(\dfrac{6}{56}\) ≤ \(\dfrac{x}{56}\) ≤ \(\dfrac{14}{56}\)
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}
\(\frac{28}{15}\times\frac{1}{4}\times3+\left(\frac{8}{15}-\frac{79}{60}\right)\times\frac{24}{47}\)
=\(\frac{7}{15}\times3+\left(\frac{32}{60}-\frac{79}{60}\right)\times\frac{24}{47}\)
=\(\frac{7}{5}+\frac{47}{60}\times\frac{24}{47}\)
=\(\frac{7}{5}+\frac{2}{5}\)
=\(\frac{5}{5}\)
=\(1\)
\(\frac{28}{15}\)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
N = 1319/4158