do cac ban biet (6/2)(1+2)=
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(x + 1) + (x + 2) + (x + 3 )+ ... + (x + 20) = 250 ( có 20 nhóm )
=> ( x + x + x +...+ x ) + ( 1 + 2 + 3 +...+ 20) = 250 ( có 20 x và 20 số hạng )
=> x . 20 + 20 . 21 : 2 = 250
=> x . 20 + 210 = 250
=> x . 20 = 250 - 210
=> x . 20 = 40
=> x = 40 : 20
x = 2
Câu 2:
Program HOC24;
var a: array[1..20] of integer;
t,i: integer;
begin
for i:=1 to 20 do
begin
write('Nhap so thu ',i,' : '); readln(a[i]);
end;
t:=0;
for i:=1 to 20 do
if a[i] mod 2<>0 then t:=t+a[i];
writeln('Tong cac so le la :',t);
readln
end.
Câu 1:
Program HOC24;
var a: array[1..10] of integer;
min,max,i: integer;
begin
for i:=1 to 10 do
begin
write('Nhap so thu ',i,' : '); readln(a[i]);
end;
min:=32000; max:=0;
for i:=1 to 10 do
begin
if max<a[i] then max:=a[i];
if min>a[i] then min:=a[i];
end;
writeln('So lon nhat la :',max);
write('So nho nhat la : ',min);
readln
end.
Câu 2:
a: Ư(4)={1;2;4}
Ư(6)={1;2;3;6}
ƯC(4;6)={1;2}
b: B(4)={0;4;8;...}
B(6)={0;6;12;18;...}
BC(4;6)={0;12;24;...}
Theo bài ra ta có:
|x+\(\frac{1}{2}\)|\(\ge\)0
|x+\(\frac{1}{6}\)|\(\ge\)0
............................
|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|\(\ge\)0
\(\Rightarrow\)11.x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
\(\Rightarrow\)x dương.
Khi đó:|x+\(\frac{1}{2}\)|+|x+\(\frac{1}{6}\)|+...+|x+\(\frac{1}{110}\)|=11.x
\(\Rightarrow\)x+\(\frac{1}{2}\)+x+\(\frac{1}{6}\)+...+x+\(\frac{1}{110}\)=11.x
\(\Rightarrow\)27.x+\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=11x
\(\Rightarrow\)\(\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)=-16x
\(\Rightarrow\)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)=-16x
\(\Rightarrow\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{11}\)=-16x
\(\Rightarrow\)\(\frac{10}{-176}=x\)
Vậy \(x=\frac{10}{-176}\).
do cac ban biet (6/2)(1+2)=