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a) \(A=\frac{135}{135.136-1}\) và \(B=\frac{136}{136.137-1}\)
\(A=\frac{1}{136-1}=\frac{1}{135}\) \(B=\frac{1}{137-1}=\frac{1}{136}\)
Vì \(\frac{1}{136}\)< \(\frac{1}{135}\)nên A > B.
a, A = \(\frac{136-1}{\left(136-1\right)136-1}\) = \(\frac{136-1}{136^2-136-1}\) B=\(\frac{136}{136\left(136+1\right)-1}\)=\(\frac{136}{136^2+136-1}\)
x=136, A-B =\(\frac{x-1}{x^2-x-1}\)-\(\frac{x}{x^2+x-1}\) =\(\frac{x^3+x^2-x-x^2-x+1-x^3+x^2+x}{\left(x^2-1\right)^2-x^2}\)=\(\frac{x^2-x+2}{\left(x^2-1\right)^2-x^2}\)<0
=> A<B
b,A = \(\frac{456-333}{456}\)= 1-333/456 B=\(\frac{789-333}{789}\)= 1-333/789
=> A>B
c, 3/14<3/13<3/12<3/11<3/10 <2/5
M = 3/10+3/11+3/12+3/13+3/14 < 2/5 x5 = 2= N
\(\left(2018-\frac{2}{135}+\frac{1}{50}\right)-\left(1-\frac{7}{135}+\frac{4}{50}\right)-\left(5+\frac{5}{135}+\frac{3}{50}\right)\)
\(=2018-\frac{2}{135}+\frac{1}{50}-1+\frac{7}{135}-\frac{4}{50}-5-\frac{5}{135}-\frac{3}{50}\)
\(=2012-\frac{6}{50}\)
\(a)\) 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
,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 \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
Đặt A = \(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2015!}\)
A < \(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2014.2015}\)
A < \(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2014}-\frac{1}{2015}\)
A < \(2-\frac{1}{2015}\)< 2 < \(2\left(\frac{135^2+136}{136^2-135}\right)\)
=> A < \(2\left(\frac{135^2+136}{136^2-135}\right)\)(Đpcm)