chứng tỏ rằng:
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2015}\right)<1\)
giúp mk nha các bn , thanks !!!!
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)....\left(1+\frac{1}{2^{50}}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.............+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có :
\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)
\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)
\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)
\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)
Mà :
\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)
Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế )
\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 )
\(\Rightarrow\)\(A>3\) ( điều phải chứng minh )
Vậy \(A>3\)
Chúc đệ học tốt ~
c,
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
vì \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.............................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
bt lm mỗi một câu :v
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)...\left(1+\frac{2}{50}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.........+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
,mình sửa lại đề:
\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)
xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015
=\(\frac{2013}{2013}\)
=\(1\)
vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
chứng tỏ \(\frac{10^{2016}+2^3}{9}\) là số tự nhiên
So sánh A=\(\left(1+\frac{1}{2016}\right)\left(1+\frac{1}{2016^2}\right)\left(1+\frac{1}{2016^3}\right)...\left(1+\frac{1}{2016^{2017}}\right)\)
\(B=\frac{2016^2-1}{2015^2-1}\)
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)
\(10^{2016}\)= 1000...00(mình ko cần biết cso bao nhiêu cx 0, nó là bài đánh lừa nhá bn)
\(2^3\)= 8
\(10^{2016}\) + 8= 10000...08
có 1+0+0+...+0+8=9. vậy số này chia hết cho 9
mà như bạn thấy số này là số dương nên số đó là số tự nhiên nhá
tinh \(G=\frac{\left(1+\frac{2015}{1}\right)+\left(1+\frac{2015}{2}\right)+...+\left(1+\frac{2015}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)+....+\left(1+\frac{1000}{2015}\right)}\)
với n thuộc N* hãy chứng tỏ rằng :
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
bài tập:
cho B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{2014}+\left(\frac{1}{2}\right)^{2015}\) . chứng minh rằng: B<1
giúp mình với nha........
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
Cho đẳng thức
\(x.\left(x+1\right).\left(x+2\right).\left(x+3\right)...\left(x+2016\right)=2016.\)
Chứng tỏ rằng x < \(\frac{1}{2015!}\)
khó zay . mik ko làm dược k cho mik ik miik kb cho
tính
A=\(\left(\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}\right)\left(1+\frac{1}{2}+...+\frac{1}{2015}\right)\left(1+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)\)