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a,\(\frac{2015.2016+2015-1}{2014+2015.2016}=\frac{2015.2016+2014}{2014+2015.2016}=1\)\(1\)
b,\(=1-\frac{1}{5}+\frac{1}{5}...-\frac{1}{2011}+\frac{1}{2011}-\frac{1}{2015}=1-\frac{1}{2015}=\frac{2014}{2015}\)
c,\(=\frac{12}{35}+\frac{12}{35}+\frac{12}{35}+\frac{12}{35}=\frac{12}{35}.4=\frac{48}{35}\)
a) Vì \(\frac{87}{39}>1\)
\(\frac{2015}{2017}< 1\)
\(\Rightarrow\frac{87}{39}>\frac{2015}{2017}\)
\(\frac{n}{n+1}\)và \(\frac{n+1}{n+3}\)
\(\Rightarrow\frac{n}{n+1}=\frac{n\cdot\left(n+3\right)}{\left(n+1\right)\left(n+3\right)}\)
\(\Rightarrow\frac{n+1}{n+3}=\frac{\left(n+1\right)^2}{\left(n+3\right)\left(n+1\right)}\)
\(\Rightarrow n\cdot\left(n+3\right)=n^2+3n\)
\(\Rightarrow\left(n+1\right)^2=n^2+2n+1\)
Dấu bằng chỉ xảy ra khi n = 1
Còn với mọi trường hợp n > 1 thì
\(\frac{n}{n+1}>\frac{n+1}{n+3};n^2+3n>n^2+2n+1\)
Gọi biểu thức trên là A
Ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}+\frac{1}{256}-\frac{1}{256}\)
\(2A=1+A-\frac{1}{256}\)
\(2A=A+1-\frac{1}{256}\)
\(2A-A=\frac{255}{256}\)
\(A=\frac{255}{256}\)
Gọi \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right]-\left[\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right]\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^8}\)
\(A=1-\frac{1}{2^8}=1-\frac{1}{256}=\frac{255}{256}\)
a, \(57< 58\Rightarrow\)\(\frac{46}{57}>\frac{46}{58}\)
b,\(1-\frac{367}{368}=\frac{1}{368}\) \(1-\frac{376}{377}=\frac{1}{377}\)
Mà \(\frac{1}{368}>\frac{1}{377}\Rightarrow1-\frac{367}{368}>1-\frac{376}{377}\)
\(\Rightarrow\frac{367}{368}< \frac{376}{377}\)
c, \(\frac{27}{26}-1=\frac{1}{26}\) \(\frac{38}{37}-1=\frac{1}{37}\)
Mà\(\frac{1}{26}>\frac{1}{37}\)\(\Rightarrow\frac{27}{26}-1>\frac{38}{37}-1\)
\(\Rightarrow\frac{27}{26}>\frac{38}{37}\)
TK NHA!
a, \(\frac{46}{57}\)> \(\frac{46}{58}\)
b, \(\frac{367}{368}\)< \(\frac{376}{377}\)
c, \(\frac{27}{26}\)<\(\frac{38}{37}\)
1 - 67/77 = 10/77 ; 1 - 73/83 = 10/83
Vì 10/77 > 10/83 nên 67/77 > 73/83
câu1:
67/77<73/83
câu2:
456/461>123/128
câu3:
2003.2004-1/2003.2004<2004.2005-1/2004.2005
câu1:
67/77<73/83
câu2:
456/461>123/128
câu3:
2003.2004-1/2003.2004<2004.2005-1/2004.2005
\(1-\frac{37}{39}=\frac{2}{39}\)
\(1-\frac{2015}{2017}=\frac{2}{2017}\)
\(\frac{2}{39}>\frac{2}{2017}\Leftrightarrow\frac{37}{39}< \frac{2015}{2017}\)
1 - 37 / 39 = 2 / 39
1 - 2015 / 2017 = 2 / 2017
2 / 39 > 2 / 2017
Nên 37 / 39 < 2015 / 2017