1+1+10=
10+10+10+10+10+10+10+10+10+10+10+10+10+10+10+1+1+1+1=?
So Sánh: A=1/(10^-1)-10/(10^0)+1/(10^1)+1/(10^2)......+1/(10^2010) và B=1000
ai chat nhìu thì kt bn với mình nha
có ai biết 1+10-10+1-10+1-10+1-10+1=........
-27 nhá bạn
so sánh 10 mũ 11-1/10 mũ 12-1 va 10 mu 10-1/10 mu 10 +1/10 mu 11-1
Số ?
2 = 1 + ... 6 = 2 + ... 8 = ...+ 3 10 = 8 + ....
3 = 1 + ... 6 =...+ 3 8 = 4 + .... 10 = ...+ 3
4 = ...+ 1 7 = 1 + ... 9 = ...+ 1 10 = 6 + ...
4 = 2 + ... 7 = ...+ 2 9 = ...+ 3 10 = ...+ 5
5 = ...+ 1 7 = 4 + .... 9 = 7 +.... 10 = 10 + ...
5 = 3 +.... 8 = ...+ 1 9 = 5 + ... 10 = 0 + .....
6 = ...+ 1 8 = 6 + ... 10 = ...+ 1 1 = 1 + ....
2 = 1 + 1 6 = 2 + 4 8 = 5 + 3 10 = 8 + 2
3 = 1 + 2 6 = 3 + 3 8 = 4 + 4 10 = 7 + 3
4 = 3 + 1 7 = 1 + 6 9 = 8 + 1 10 = 6 + 4
4 = 2 + 2 7 = 5 + 2 9 = 6+ 3 10 = 5 + 5
5 = 4 + 1 7 = 4 + 3 9 = 7 + 2 10 = 10 + 0
5 = 3 + 2 8 = 7 + 1 9 = 5 + 4 10 = 0 + 10
6 = 5 + 1 8 = 6 + 2 10 = 9 + 1 1 = 1 + 0
Tiếng việt khó .
so sánh:A=10^10+1/10^11+1 và B=10^9+1/10^10+1
10^10+1/10^3+1 và 10^9+1/10^8+1
\(10^{10}+\frac{1}{10}^{10}=10^{10}\)
\(10^9+\frac{1}{10}^8+1=10^9+1\)
\(10^{10}>10^9+1\)
so sánh
a) 1/2^2+1/2^3+...1/2^2014 và 1
b)A=10^11-1/10^12-1 và B=10^10+1/10^11+1
Giải:
a) Gọi dãy đó là A, ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\)
\(2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\)
\(2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^{2014}}\)
Vì \(\dfrac{1}{2}< 1;\dfrac{1}{2^{2014}}< 1\) nên \(\dfrac{1}{2}-\dfrac{1}{2^{2014}}< 1\)
\(\Rightarrow A< 1\)
b) \(A=\dfrac{10^{11}-1}{10^{12}-1}\) và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
Ta có:
\(A=\dfrac{10^{11}-1}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-10}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-1+9}{10^{12}-1}\)
\(10A=1+\dfrac{9}{10^{12}-1}\)
Tương tự:
\(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+1+9}{10^{11}+1}\)
\(10B=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\) nên \(10A< 10B\)
\(\Rightarrow A< B\)
\(\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}+\dfrac{1}{10}x7+\dfrac{1}{2}\)
So Sánh : \(\dfrac{10^{11}-1}{10^{12}-1}\)và\(\dfrac{10^{10}+1}{10^{11}+1}\)
Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
Vì \(10^{11}-1< 10^{12}-1\)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)