1+10=
10+10+10+10+10+10+10+10+10+10+10+10+10+10+10+1+1+1+1=?
so sánh 10 mũ 11-1/10 mũ 12-1 va 10 mu 10-1/10 mu 10 +1/10 mu 11-1
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: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
\(10^{10}+\frac{1}{10}^{10}=10^{10}\)
\(10^9+\frac{1}{10}^8+1=10^9+1\)
\(10^{10}>10^9+1\)
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 : \(\dfrac{10^{11}-1}{10^{12}-1}\)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 \(10^{11}-1< 10^{12}-1\)
\(\Rightarrow\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10^{10}+1}{10^{11}+1}\)
1. So sánh
a) A=\(\dfrac{10^{15}.11}{10^{16}+1}\) với B=\(\dfrac{10^{16}+10}{10^{16}+1}\)
b) C+\(\dfrac{10^{10}+1}{10^{10}-1}\) với D=\(\dfrac{10^{10}-1}{10^{13}-3}\)
a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)
Vì \(10^{16}+10^{15}>10^{16}+10\)
\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)
Hay A>B
b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)
\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)
Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)
\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)
Hay C > D
10^8 +1 / 10^9 + 1 và 10^9 +1/ 10^10 +1
???? thiếu đề.....
bạn vào sửa nội dung nhak
~~~
Ta chứng minh bài toán phụ:
Nếu \(\frac{a}{b}< 1\)thì \(\frac{a}{b}< \frac{a+c}{b+c}\)
Ta có: \(a< b\)
\(\Rightarrow ac< bc\)
\(\Rightarrow ac+ba< bc+ba\)
\(\Rightarrow a.\left(b+c\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)
đpcm
Áp dụng:
\(\frac{10^9+1}{10^{10}+1}< \frac{10^9+1+9}{10^{10}+1+9}=\frac{10^9+10}{10^{10}+10}=\frac{10.\left(10^8+1\right)}{10.\left(10^9+1\right)}=\frac{10^8+1}{10^9+1}\)
Vậy \(\frac{10^9+1}{10^{10}+1}< \frac{10^8+1}{10^9+1}\)
Tham khảo nhé~