Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{10^{2012}+1}{10^{2013}+1}\)
\(10A=\frac{10\cdot\left[10^{2012}+1\right]}{10^{2013}+1}=\frac{10^{2013}+10}{10^{2013}+1}=\frac{10^{2013}+1+9}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}\)
\(B=\frac{10^{2013}+1}{10^{2014}+1}\)
\(10B=\frac{10\cdot\left[10^{2013}+1\right]}{10^{2014}+1}=\frac{10^{2014}+10}{10^{2014}+1}=\frac{10^{2014}+1+9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
Mà \(1+\frac{9}{10^{2013}+1}>1+\frac{9}{10^{2014}+1}\)
Nên \(10A>10B\)
Hay \(A>B\)
Vậy : A > B
b, 2000A = \(\frac{2000\left(2000^{2015}+1\right)}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+2000}{2000^{2016}+1}\)
= \(\frac{\left(2000^{2016}+1\right)+1999}{2000^{2016}+1}\)
= \(\frac{2000^{2016}+1}{2000^{2016}+1}\) + \(\frac{1999}{2000^{2016}+1}\)
= 1 + \(\frac{1999}{2000^{2016}+1}\)
2000B = \(\frac{2000\left(2000^{2014}+1\right)}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+2000}{2000^{2015}+1}\)
= \(\frac{\left(2000^{2015}+1\right)+1999}{2000^{2015}+1}\)
= \(\frac{2000^{2015}+1}{2000^{2015}+1}\) + \(\frac{1999}{2000^{2015}+1}\)
= 1 + \(\frac{1999}{2000^{2015}+1}\)
So sanh
câu b tiếp
So sánh 2000A với 2000B
Vì \(\frac{1999}{2000^{2016}+1}\) < \(\frac{1999}{2000^{2015}+1}\)
→ 2000A< 2000B
→ A<B
TA có :
A = \(\frac{10^{2012}-2}{10^{2013}-1}\)=> 10A = \(1-\frac{19}{10^{2013}-1}\)
B = \(\frac{10^{2013}-2}{10^{2014}-1}\)=> 10B = 1 - \(\frac{19}{10^{2014}-1}\)
Vì \(1-\frac{19}{10^{2013}-1}\)< 1 - \(\frac{19}{10^{2014}-1}\)hay 10A < 10B => A < B
Vậy A < B
vì B<1 => \(B=\frac{10^{2013}+1}{10^{2014}+1}< \frac{10^{2013}+1+9}{10^{2014}+1+9}=\)\(\frac{10^{2013}+10}{10^{2014}+10}=\frac{10\left(10^{2012}+1\right)}{10\left(10^{2013}+1\right)}\)\(=\frac{10^{2012}+1}{10^{2013}+1}=A\)
\(\Rightarrow A>B\)
\(\frac{10^{2012}+1}{10^{2013}+1}=\frac{\left(10^{2012}+1\right)\cdot10}{\left(10^{2013}+1\right)\cdot10}=\frac{10^{2013}+10}{\left(10^{2013}+1\right)\cdot10}=\frac{10^{2013}+1+9}{\left(10^{2013}+1\right)\cdot10}=\frac{10^{2013}+1}{\left(10^{2013}+1\right)\cdot10}+\frac{9}{\left(10^{2013}+1\right)\cdot10}=\frac{1}{10}+\frac{9}{\left(10^{2013}+1\right)\cdot10}\left(1\right)\)
\(\frac{10^{2013}+1}{10^{2014}+1}=\frac{\left(10^{2013}+1\right)\cdot10}{\left(10^{2014}+1\right)\cdot10}=\frac{10^{2014}+10}{\left(10^{2014}+1\right)\cdot10}=\frac{10^{2014}+1+9}{\left(10^{2014}+1\right)\cdot10}=\frac{10^{2014}+1}{\left(10^{2014}+1\right)\cdot10}+\frac{9}{\left(10^{2014}+1\right)\cdot10}=\frac{1}{10}+\frac{9}{\left(10^{2014}+1\right)\cdot10}\left(2\right)\)Từ (1)(2) => A > B
Vì \(\frac{10^{2014}+1}{10^{2015}+1}< 1\Rightarrow B=\frac{10^{2014}+1}{10^{2015}+1}< \frac{10^{2014}+1+9}{10^{2015}+1+9}\)
\(\Rightarrow B< \frac{10^{2014}+10}{10^{2015}+10}\)
\(\Rightarrow B< \frac{10\left(10^{2013}+1\right)}{10\left(10^{2014}+1\right)}\)
\(\Rightarrow B< \frac{10^{2013}+1}{10^{2014}+1}\)
\(\Rightarrow B< A\)
Vậy A > B
có :
\(B=\frac{10^{2015}+1}{10^{2014}+1}>1\)
\(\Rightarrow\frac{10^{2015}+1}{10^{2014}+1}>\frac{10^{2015}+1+9}{10^{2014}+1+9}\) \(=\frac{10^{2015}+10}{10^{2014}+10}=\frac{10.\left(10^{2014}+1\right)}{10.\left(10^{2013}+1\right)}\)
\(=\frac{10^{2014}+1}{10^{2013}+1}=A\)
\(\Rightarrow B>A\)
Vậy B > A
k cho mk nhé
\(\Rightarrow10A=10.\left(\frac{10^{2012}+1}{10^{2013}+1}\right)=\frac{10^{2013}+10}{10^{2013}+1}=\frac{10^{2013}+1+9}{10^{2013}+1}=1+\frac{9}{10^{2013}+1}\)
\(\Rightarrow10B=10.\left(\frac{10^{2013}+1}{10^{2014}+1}\right)=\frac{10^{2014}+10}{10^{2014}+1}=\frac{10^{2014}+1+9}{10^{2014}+1}=1+\frac{9}{10^{2014}+1}\)
Ta có: 1 = 1; 9 = 9
Mà \(10^{2013}+1<10^{2014}+1\)
=> \(\frac{9}{10^{2013}+1}>\frac{9}{10^{2014}+1}\)
=> \(1+\frac{9}{10^{2013}+1}>1+\frac{9}{10^{2014}+1}\text{ hay }10A>10B\)
=> \(A>B\).
a)Ta áp dụng tính chất sau:
Nếu a<b=>a/b<(a+k)/(b+k) (k thuộc N*)
Vì 1013+1<1014+1=>B=1013+1/1014+1<1013+1+9/1014+1+9
=>B<1013+10/1014+10
=>B<10.(1012+1)/10.(1013+1)
=>B<1012+1/1013+1=A
=>B<A
b)Ta áp dụng tính chất sau:
Nếu a>b=>a/b>(a+k)/(b+k) (k thuộc N*)
Vì 102015+1>102014+1=>B=102015+1/102014+1>102015+1+99/102014+1+99
=>B>102015+100/102014+100
=>B>100.(102013+1)/100.(102012+1)
=>B>102013+1/102012+1=A
=>B>A
Mình làm cho câu đầu tiên thôi, câu thứ hai cũng tương tự nha:
Ta có:
A.10 = \(\frac{10^{12}+10}{10^{12}+1}\) B.10 = \(\frac{10^{14}+10}{10^{14}+1}\)
=>A.10 = \(\frac{10^{12}+1+9}{10^{12}+1}\) =>B.10 = \(\frac{10^{14}+1+9}{10^{14}+1}\)
=>A.10 = 1 + \(\frac{9}{10^{12}+1}\) =>B.10 = 1 + \(\frac{9}{10^{14}+1}\)
=>A.10 > B.10
=>A > B
Vậy A > B