Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
Đặt \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{n}{3^n}\)
\(3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{n}{3^{n-1}}\)
\(\Rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}-\dfrac{n}{3^n}< 1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}\)
Đặt \(B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{n-1}}\)
Tương tự ta được \(2B=3-\dfrac{1}{3^{n-1}}< 3\)
\(\Rightarrow B< \dfrac{3}{2}\Rightarrow2A< \dfrac{3}{2}\Rightarrow A< \dfrac{3}{4}\)(đpcm)
BonkingTrần Trung Nguyên làm giùm bài này luôn đi
\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
a, vì m>n
=> m+7>n+7
b, vì m>n
=> -2m<-2n
=>-2m-8<-2n-8
c, vì m>n
=>m+1>n+1
mà m+3>m+1
=>m+3>n+1
phần d,e,f máy mình cùi nên không hiện ra phép tính. sr nhiều
m>n
a) m+7 và m+7
ta có : m>n
=> m+7 > n+7
b) -2m+8 và -2n+8
ta có : m>n
=> -2m > -2n
=> -2m+8 > -2n+8
c) m+3 và m+1
ta có : 3 >1
=> m+3 > m+1
d) \(\dfrac{1}{2}\) \(\left(m-\dfrac{1}{4}\right)\)và\(\dfrac{1}{2}\)\(\left(n-\dfrac{1}{4}\right)\)
ta có: m > n
=> \(m-\dfrac{1}{4}\) > \(n-\dfrac{1}{4}\)
=>\(\dfrac{1}{2}\left(m-\dfrac{1}{4}\right)\)>\(\dfrac{1}{2}\left(n-\dfrac{1}{4}\right)\)
e) \(\dfrac{4}{5}-6\)m và \(\dfrac{4}{5}-6n\)
ta có : m > n
=> -6m > -6n
=> \(\dfrac{4}{5}-6m>\dfrac{4}{5}-6n\)
f) \(-3\left(m+4\right)+\dfrac{1}{2}\) và \(-3\left(n+4\right)+\dfrac{1}{2}\)
ta có : m > n
=> m=4 > n+4
=> -3(m+4) > -3(m+4)
=>\(-3\left(m+4\right)+\dfrac{1}{2}>-3\left(n+4\right)+\dfrac{1}{2}\)
a)Nhận xét
\(\dfrac{n^3+1}{n^3-1}=\dfrac{\left(n+1\right)\left(n^2-n+1\right)}{\left(n-1\right)\left(n^2+n+1\right)}=\dfrac{\left(n+1\right)\left[\left(n-0,5\right)^2+0;75\right]}{\left(n-1\right)\left[\left(n+0,5\right)^2+0,75\right]}\)
Áp dụng công thức trên:
\(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}....\dfrac{9^3+1}{9^3-1}\)
\(=\dfrac{\left(2+1\right)\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right)\left[\left(2+0,5\right)^2+0,75\right]}.\dfrac{\left(3+1\right)\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right)\left[\left(3+0,5\right)^2+0,75\right]}...\dfrac{\left(9+1\right)\left[\left(9-0,5\right)^2+0,75\right]}{\left(9-1\right)\left[\left(9+0,5\right)^2+0,75\right]}\)
\(=\dfrac{3\left(1,5^2+0,75\right)}{\left(2,5^2+0,75\right)}.\dfrac{4\left(2,5^2+0,75\right)}{2\left(3,5^2+0,75\right)}...\dfrac{10\left(8,5^2+0,75\right)}{8\left(9,5^2+0,75\right)}\)
\(=\dfrac{3.4....10}{1.2.....8}.\dfrac{1,5^2+0,75}{9,5^2+0,75}\)
\(=\dfrac{9.10}{2}.\dfrac{3}{91}\)
\(=\dfrac{3}{2}.\dfrac{90}{91}< \dfrac{3}{2}\)
\(\Rightarrowđpcm\)
b) Làm tương tự
ta có \(\dfrac{1}{3^3}< \dfrac{1}{3^3-3}\)
\(\dfrac{1}{4^3}< \dfrac{1}{4^3-4}\)
...............
\(\dfrac{1}{n^3}< \dfrac{1}{n^3-n}\)
=> \(\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+....+\dfrac{1}{n^3}< \dfrac{1}{3^3-3}+\dfrac{1}{4^3-4}+....+\dfrac{1}{n^3-n}\)=>\(B< \dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)đặt \(C=\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)
C=\(\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+.....+\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\)C=\(\dfrac{1}{6}-\dfrac{1}{n\left(n+1\right)}\)
=> C<\(\dfrac{1}{6}\)
mà\(\dfrac{1}{6}< \dfrac{1}{4}\)
=> C<\(\dfrac{1}{4}\)
ta lại có B<C
=> B<\(\dfrac{1}{4}\) (đpcm)