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vì bài dài quá nên mình làm từng bài 1 nhé
1. Ta thấy : \(\frac{1}{n^3}< \frac{1}{n^3-n}=\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Do đó :
\(B< \frac{1}{2}.\left[\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]< \frac{1}{2}.\frac{1}{6}=\frac{1}{12}\)
2.
Nhận xét : \(1+\frac{1}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n\left(n+2\right)}\)
Do đó :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{\left(n+1\right)^2}{n\left(n+2\right)}=\frac{2.3...\left(n+1\right)}{1.2...n}.\frac{2.3...\left(n+1\right)}{3.4...\left(n+2\right)}=\frac{n+1}{1}.\frac{2}{n+2}< 2\)
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
a; A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + ... + \(\dfrac{1}{\left(2n\right)^2}\)
A = \(\dfrac{1}{2^2}\).(\(\dfrac{1}{1^2}\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + ... + \(\dfrac{1}{n^2}\))
A = \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ... + \(\dfrac{1}{n.n}\))
Vì \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\); \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\); ...; \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right)n}\)
nên A < \(\dfrac{1}{4}\).(\(\dfrac{1}{1}\) + \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{\left(n-1\right)n}\))
A < \(\dfrac{1}{4.}\)(1 + \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(1 + 1 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{4}\).(2 - \(\dfrac{1}{n}\))
A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\) (đpcm)
Từ "lạc trôi" có nghĩa là gì trong câu:
"Mây bềnh bồng lạc trôi/mượt mà như tuổi ngọc."
1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
Mình xem phép làm câu 1 ạ.
Đề là?
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\)(1)
Chứng minh tương đương
\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge4\)<=> 12ac - 9bc - 9ab + 6b2 \(\le\)0 ( quy đồng ) (2)
Từ (1) <=> 2ac = ab + bc Thay vào (2) <=> 6ab + 6bc - 9bc - 9ab + 6b2 \(\le\)0
<=> a + c \(\ge\)2b
Từ (1) => \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)
=> a + c \(\ge\)2b đúng => BĐT ban đầu đúng
Dấu "=" xảy ra <=> a = c = b
Ta có: \(\left(2a+1\right)^2>\left(2a+1\right)^2-1\)
\(\Leftrightarrow\left(2a+1\right)^2>2a.\left(2a+2\right)\)
\(\Rightarrow\frac{1}{\left(2a+1\right)^2}< \frac{1}{2a.\left(2a+2\right)}\)(*)
ĐẶT \(A=\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{\left(2a+1\right)^2}\)
Áp dụng (*), ta có:
\(A< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2a.\left(2a+2\right)}\)
\(\Leftrightarrow A< \frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2a.\left(2a+2\right)}\right)\)
\(\Leftrightarrow A< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2a}-\frac{1}{2a+2}\right)\)
\(\Leftrightarrow A< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{2a+2}\right)\)
\(\Leftrightarrow A< \frac{1}{4}-\frac{1}{4a+4}< \frac{1}{4}\)
Vậy ..........
Có : 3^2 = 2.4+1
5^2 = 4.6 +1
..........
(2a+1)^2 = 2a.(2a+2)+1
=> VT < 1/2.4 + 1/4.6 + .... + 1/2a.(2a+2)
2VT < 2/2.4 + 2/4.6 + .... + 2/2a.(2a+2)
= 1/2 - 1/4 + 1/4 - 1/6 + ..... 1/2a - 1/2a+2 = 1/2 - 1/2a+2 < 1/2
=> VT < 1/2 (ĐPCM)