So sánh
A=\(\frac{2005^{2014}+1}{2005^{2015}+1}\)
với
B=\(\frac{2005^{2015}+1}{2005^{2016}+1}\)
K
Khách
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DV
0
SY
1 tháng 4 2016
A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1 => \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\) = \(\frac{2005^{2005}+2005}{2005^{2006}+2005}\)= \(\frac{2005.\left(2005^{2004}+1\right)}{2005.\left(2005^{2005}+1\right)}\) = \(\frac{2005^{2004}+1}{2005^{2005}+1}\) = B => A<B.
DT
4
22 tháng 7 2017
haizzz mk nhớ bài này nhìu người hỏi lắm rồi,chịu khó tìm là thấy
MV
10 tháng 8 2017
ồ, lâu h ms gặp
a,
Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)
Ta có:
\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
b,
\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)
Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)
c,
\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)
Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)
Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)
10 tháng 8 2017
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)
\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)
\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)
\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)
\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)
\(B< A\)
12 tháng 2 2017
\(\frac{100^{2015}+1}{100^{2015}+1}=1\)
\(\frac{100^{2016}+1}{100^{2016}+1}=1\)
Vì 1 = 1 nên \(\frac{100^{2015}+1}{100^{2015}+1}=\frac{100^{2016}+1}{100^{2016}+1}\)
12 tháng 2 2017
à mình nhìn nhầm đề
Mình giải nha
Đặt \(A=\frac{100^{2015}+1}{100^{2005}+1}\Rightarrow\frac{A}{100^{10}}=\frac{100^{2015}+1}{100^{2015}+100^{10}}=\frac{100^{2015}+100^{10}-999}{100^{2015}+100^{10}}=1-\frac{999}{100^{2015}+100^{10}}\)
Đặt \(B=\frac{100^{2016}+1}{100^{2006}+1}\Rightarrow\frac{B}{100^{10}}=\frac{100^{2016}+100^{10}-999}{100^{2016}+100^{10}}=1-\frac{999}{100^{2016}+100^{10}}\)
\(1-\frac{999}{100^{2015}+100^{10}}< 1-\frac{999}{100^{2016}+100^{10}}\Rightarrow A< B\)
B
23 tháng 5 2016
mỗi số hạng trong biểu thức A đều nhỏ hơn 1 mà có 15 số nên tổng A sẽ nhỏ hơn 15
23 tháng 5 2016
ta thay tong tren <1+1+1+1+1+1+1+1+1+1+1+1+1+1+1
hay tong tren be hon 15
13 tháng 6 2018
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
Ta thấy \(\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\)
Suy ra \(1+\frac{2004}{2005^{2005}+1}>1+\frac{2004}{2005^{2006}+1}\)
hay 2005B>2005A
Vậy B>A
23 tháng 5 2018
Ta có VẾ A
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005\cdot A=1+\frac{2004}{2005^{2006}+1}\)
Ta lại có Vế B :
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)
\(2005\cdot B=1+\frac{2004}{2005^{2005}+1}\)
Nhìn vào trên , suy ra A < B .
WH
23 tháng 5 2018
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{2005^{2014}+1}{2005^{2015}+1}< 1\\B=\dfrac{2005^{2015}+1}{2005^{2016}+1}< 1\end{matrix}\right.\)
\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:
\(\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2015}+1+2004}{2005^{2016}+1+2004}\)
\(=\dfrac{2005^{2015}+2005}{2005^{2016}+2005}=\dfrac{2005\left(2005^{2014}+1\right)}{2005\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)
\(\Rightarrow\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)
Vậy \(B< A\)
Hay \(A>B\)