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MSC (2,4,9,5) = 180
\(\dfrac{1}{2}=\dfrac{90}{180};\dfrac{3}{4}=\dfrac{135}{180};\dfrac{8}{9}=\dfrac{160}{180};\dfrac{8}{5}=\dfrac{288}{180}\)
Vì: \(\dfrac{90}{180}< \dfrac{135}{180}< \dfrac{160}{180}< \dfrac{288}{180}\) nên \(\dfrac{1}{2}< \dfrac{3}{4}< \dfrac{8}{9}< \dfrac{8}{5}\)
Vậy: đáp án D. `8/5` là đáp án đúng.
Bạn có thể tham khảo cách ngắn gọn hơn:D.
\(\dfrac{1}{2}< 1;\dfrac{3}{4}< 1;\dfrac{8}{9}< 1;\dfrac{8}{5}>1\)
Vì \(\dfrac{8}{5}>1\) nên đáp án D là đáp án đúng.
https://www.youtube.com/channel/UC5odkiOvzz9Rvu3HUYlL2IQ?view_as=subscriber
a) \(\dfrac{13}{15}< \dfrac{14}{15}\) ( vì 13 < 14 )
b) Ta có:
\(\dfrac{2}{3}=\dfrac{2\times3}{3\times3}=\dfrac{6}{9}\)
Vì 6 = 6 nên \(\dfrac{6}{9}=\dfrac{6}{9}\)
Vậy \(\dfrac{2}{3}=\dfrac{6}{9}\)
c) Ta có:
\(\dfrac{4}{7}=\dfrac{4\times6}{7\times6}=\dfrac{24}{42}\)
\(\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)
Vì 24 < 35 nên \(\dfrac{24}{42}< \dfrac{35}{42}\)
Vậy \(\dfrac{4}{7}< \dfrac{5}{6}\)
d) Vì 9 < 13 nên \(\dfrac{4}{9}>\dfrac{4}{13}\)
Vậy \(\dfrac{4}{9}>\dfrac{4}{13}\)
b) 1-3+5-7+9-11+......+2005-2007
=(1-3)+(5-7)+(9-11)+.....+(2005-2007)
=(-2)+(-2)+(-2)+......+(-2)
=(-2).1004
=(-2008)
c) 1+2+3-4-5-6+7+8+9-10-11-12+...+97+98+99-100-101-102
=(1+2+3-4-5-6)+(7+8+9-10-11-12)+.....+(97+98+99-100-101-102)
=(-9)+(-9)+....+(-9)
=(-9).17
=(-153)
Xin lỗi nha 2 dòng cuối mk làm sai
b)1-3+5-7+9-11+......+2005-2007
=(1-3)+(5-7)+(9-11)+....+(2005-2007)
=(-2)+(-2)+(-2)+....+(-2)
=(-2).502
=(-1004)
a,(0,8)5:(0,4)6 = (\(\dfrac{0,8}{0,4}\))5 : 0,4 = 25:0,4 = 80
b, (-25)7: 323 - \(\dfrac{6103515625}{3276}\) = - 186264,5149
c, \(\dfrac{4^2.4^3}{2^{10}}\) = \(\dfrac{4^5}{2^{10}}\) = \(\dfrac{2^{10}}{2^{10}}\) = 1
d, \(\dfrac{9^5.5^7}{45^7}\) = \(\dfrac{9^5.5^7}{9^7.5^7}\) = \(\dfrac{1}{81}\)
\(\dfrac{\left(0.8\right)^5}{\left(0.4\right)^4}\)=\(\dfrac{\left(2.0,4\right)^5}{\left(0.4^4\right)}\)=\(\dfrac{2^5.\left(0.4\right)^5}{\left(0,4\right)^4}\)=\(2^5\).\(\left(0.4\right)^1\)=12,8
b)câu b không biết có sai đề không nhưng đáp án câu b là -186264,5149
c) \(\dfrac{4^2.4^3}{2^{10}}\)=\(\dfrac{4^5}{\left(2^2\right)^5}\)=\(\dfrac{4^5}{4^5}\)=1
d)\(\dfrac{9^5.5^7}{45^7}\)=\(\dfrac{9^5.5^5.5^2}{45^7}\)=\(\dfrac{45^5.5^2}{45^7}\)=\(\dfrac{5^2}{45^2}\)=\(\left(\dfrac{5}{45}\right)^2\)=\(\left(\dfrac{1}{9}\right)^2\)=\(\dfrac{1}{81}\)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
#include <bits/stdc++.h>
using namespace std;
long long a[1000],n,i,tam,j;
int main()
{
cin>>n;
for (i=1; i<=n; i++)
cin>>a[i];
for (i=1; i<=n-1; i++)
for (j=i+1; j<=n; j++)
if (a[i]>a[j]) swap(a[i],a[j]);
for (i=1; i<=n; i++)
cout<<a[i]<<" ";
return 0;
}
C
Chọn C