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haizzz mk nhớ bài này nhìu người hỏi lắm rồi,chịu khó tìm là thấy
\(\frac{100^{2015}+1}{100^{2015}+1}=1\)
\(\frac{100^{2016}+1}{100^{2016}+1}=1\)
Vì 1 = 1 nên \(\frac{100^{2015}+1}{100^{2015}+1}=\frac{100^{2016}+1}{100^{2016}+1}\)
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 \(B=\frac{100^{2016}+1}{100^{2006}+1}\Rightarrow\frac{B}{100^{10}}=\frac{100^{2016}+100^{10}-999}{100^{2016}+100^{10}}=1-\frac{999}{100^{2016}+100^{10}}\)
\(1-\frac{999}{100^{2015}+100^{10}}< 1-\frac{999}{100^{2016}+100^{10}}\Rightarrow A< B\)
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)
\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)
\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)
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
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 .
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
\(\Rightarrow A< B\)
\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\) và \(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
So sánh A và B
\(2005A=\frac{2005^{2005}+1}{2005^{2006}+1}=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}}\) \(=\frac{2005^{2006}+2014+1}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005^{2004}+1}{2005^{2005}+1}=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(=\frac{2005^{2005}+2004+1}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Vì \(2005^{2006}+1>2005^{2005}+1\)
Nên \(1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)
Hay A < B
Vậy A < B
sửa chỗ \(\frac{2005^{2006}+2014+1}{2005^{2006}+1}\) thành \(\frac{2005^{2006}+2004+1}{2005^{2006}+1}\)nhé
Nhân a và b với 2005 ta có : 2005.a =\(\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}\)=\(\frac{2005^{2006}+2005}{2005^{2006}+1}\)= \(\frac{\left(2005^{2006}+1\right)+2004}{2005^{2006}+1}\)= \(\frac{2005^{2006}+1}{2005^{2006}+1}\)+ \(\frac{2004}{2005^{2006}+1}\)=1+\(\frac{2004}{2005^{2006}+1}\) 2005.b = \(\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}\)=\(\frac{2005^{2005}+2005}{2005^{2005}+1}\)= \(\frac{\left(2005^{2005}+1\right)+2004}{2005^{2005}+1}\)=\(\frac{2005^{2005}+1}{2005^{2005}+1}\)+ \(\frac{2004}{2005^{2005}+1}\) =1+\(\frac{2004}{2005^{2005}+1}\) Vì 2004=2004 , 2005^2005 +1 < 2005^2006 + 1 => \(\frac{2004}{2005^{2006}+1}\)< \(\frac{2004}{2005^{2005}+1}\)=> a<b Vậy A < B
B=(2005(2005^2004+1))/(2005(2005^2005+1))=(2005^2005+2005)/(2005^2006+2005)
Có 1-A=(2005^2006-2005^2005)/(2005^2006+1)
1-B=(2005^2006-2005^2005)/(2005^2006+2005)
suy ra 1-A>1-B.Suy ra A <B
\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
vì 20052006+1>20052005+1
\(\Rightarrow\frac{4}{2005^{2006}+1}< \frac{4}{2005^{2005}+1}\)
\(\Rightarrow1+\frac{4}{2005^{2006}+1}< 1+\frac{4}{2005^{2005}+1}\)
=>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 \(A>B\)