Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu hỏi của thục hà - Toán lớp 6 - Học toán với OnlineMath
Em tham khảo nhé!
Đề sai hả
\(P=\frac{12}{1.4.7}+\frac{12}{4.7.10}+...+\frac{12}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{6}{1.4.7}+\frac{6}{4.7.10}+...+\frac{6}{54.57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{4.7}+\frac{1}{4.7}-\frac{1}{7.10}+...+\frac{1}{54.57}-\frac{1}{57.60}\)
\(\Rightarrow\frac{1}{2}P=\frac{1}{1.4}-\frac{1}{57.60}< \frac{1}{4}\)
\(\Rightarrow P< \frac{1}{4}.2=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét với mọi n > 2 , ta có \(\frac{n}{n+2}< \frac{n-1}{n}\) (vì \(n^2< n^2+n-2\))
Áp dụng : \(A=\frac{1}{3}.\frac{4}{6}.\frac{7}{9}.\frac{10}{12}...\frac{208}{210}< \frac{1}{3}.\frac{3}{4}.\frac{6}{7}.\frac{9}{10}...\frac{207}{208}\)
Suy ra : \(A^2< \frac{1.4.7.10...208}{3.6.9.12...210}.\frac{1.3.6.9...207}{3.4.7.10...208}=\frac{1}{210}.\frac{1}{3}=\frac{1}{630}< \frac{1}{625}=\left(\frac{1}{25}\right)^2\)
Do đó \(A< \frac{1}{25}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(B=\dfrac{10^{12}+1}{10^{12}+1}=1\)
+) Xét \(n>12\Rightarrow A>1=B\)
+) Xét \(n< 12\Rightarrow A< B=1\)
Vậy...
b, \(\overline{abc}-\overline{deg}⋮7\)
\(\Rightarrow\left\{{}\begin{matrix}\overline{abc}⋮7\\\overline{deg}⋮7\end{matrix}\right.\)
Ta có: \(\overline{abcdeg}=1000\overline{abc}+\overline{deg}⋮7\) ( do \(\left(1000;7\right)=1\) )
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{12}{1.4.7}+\dfrac{12}{4.7.10}+\dfrac{12}{7.10.13}+...+\dfrac{12}{54.57.60}\)
\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{4.7}+\dfrac{1}{4.7}-\dfrac{1}{7.10}+\dfrac{1}{7.10}-\dfrac{1}{10.13}+...+\dfrac{1}{54.57}-\dfrac{1}{57.60}\right)\)\(=2\left(\dfrac{1}{1.4}-\dfrac{1}{57.60}\right)\)
\(=2\left(\dfrac{1}{4}-\dfrac{1}{57.60}\right)=\dfrac{1}{2}-\dfrac{1}{2.57.60}< \dfrac{1}{2}\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B1\)
\(\frac{3}{4}^{-2}=\frac{16}{9}\)
B 3
\(A=2^{13}\times3^{19}\)
mk mới lớp 6 thui bạn ạ
////????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????không biết