\(A=\frac{10^{2001}+1}{10^{2002}+1}\)
\(B=\frac{10^{2002}+1}{10^{2003}+1}\)
Hãy so sánh A và B
K
Khách
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NK
19 tháng 2 2016
\(A=\frac{10^{2001}+1}{10^{2002}+1}=\frac{\left(10^{2001}+1\right)\left(10^{2003}+1\right)}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}=\frac{10^{4004}+10^{2001}+10^{2003}+1}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}\)
\(B=\frac{10^{2002}+1}{10^{2003}+1}=\frac{\left(10^{2002}+1\right)\left(10^{2002}+1\right)}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}=\frac{10^{4004}+2.10^{2002}+1}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}\)
Vì 102001 + 102003 < 2.102002 nên A < B
AP
A = \(\frac{10^{2001}+1}{10^{2002}+1}\). B = \(\frac{10^{2002}+1}{10^{2003}+1}\). Hãy so sánh A và B
4
14 tháng 5 2016
10^2002/10^2003<1 =>B =10^2002+1/10^2003+1<10^2002+1+9/10^2003+1+9
=10^2001+10/10^2003+10
=10.(10^2001+1)/10.(10^2002+1)
=10^2001/10^2002=A
Vậy A< B
HP
14 tháng 5 2016
\(A=\frac{10^{2001}+1}{10^{2002}+1}\Rightarrow10A=\frac{10.\left(10^{2001}+1\right)}{10^{2002}+1}=\frac{10^{2002}+10}{10^{2002}+1}\)
\(10A=\frac{10^{2002}+1+9}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)
\(B=\frac{10^{2002}+1}{10^{2003}+1}\Rightarrow10B=\frac{10.\left(10^{2002}+1\right)}{10^{2003}+1}=\frac{10^{2003}+10}{10^{2003}+1}\)
\(10B=\frac{10^{2003}+1+9}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)
Vì \(10^{2002}+1<10^{2003}+1\Rightarrow\frac{9}{10^{2002}+1}>\frac{9}{10^{2003}+1}\Rightarrow10A>10B\Rightarrow A>B\)
DH
1
12 tháng 4 2019
Ta c/m bài toán phụ:
Giả sử a<b (a,b\(\in\)N; b\(\ne\)0)
So sánh \(\frac{a}{b}\) với \(\frac{a+m}{b+m}\) (m\(\in\)N*)
Có: \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b\left(b+m\right)}\)
\(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b\left(b+m\right)}\)
Vì a<b \(\Rightarrow\) am<bm (m\(\in\)N*) \(\Rightarrow\) ab+am<ab+bm
\(\Rightarrow\frac{ab+am}{b\left(b+m\right)}< \frac{ab+bm}{b\left(b+m\right)}\) hay \(\frac{a}{b}< \frac{a+m}{b+m}\)
Áp dụng bài toán trên ta có:
\(B=\frac{10^{2002}+1}{10^{2003}+1}< \frac{10^{2002}+1+9}{10^{2003}+1+9}=\frac{10^{2002}+10}{10^{2003}+10}=\frac{10\left(10^{2001}+1\right)}{10\left(10^{2002}+1\right)}=\frac{10^{2001}+1}{10^{2002}+1}=A\)
\(\Rightarrow B< A\)
Vậy B<A
Y
0
ND
1
TL
2 tháng 4 2017
Ta có: 10 *(10^2001+1)/10^2002+1 = 10^2002+10/10^2002+1 = (10^2002+1)+9/10^2002+1 = 1+9/10^2002+1
10*(10^2002+1)/10^2003+1 = 10^2003+10/10^2003+1 = (10^2003+1)+9/10^2003+1 = 1+9/10^2003+1
Vì 9/10^2002+1>9/10^2003+1 nên 1+9/10^2002+1>1+9/10^2003+1
Vậy: 10^2001+1/10^2002+1>10^2002+1/10^2003+1
VN
4
16 tháng 3 2019
Tham khảo:Câu hỏi của Trần Trí Trung - Toán lớp 6 - Học toán với OnlineMath
AT
0
ta thấy:
\(B< 1\Rightarrow B< \frac{10^{2002}+1+9}{10^{2003}+1+9}=\frac{10^{2002}+10}{10^{2003}+10}=\frac{10\left(10^{2001}+1\right)}{10\left(10^{2002}+1\right)}=\frac{10^{2001}+1}{10^{2002}+1}=A\)
=>B<A
vậy.......
Ta có:
\(A=\frac{10^{2001}+1}{10^{2002}+1}\Rightarrow10A=\frac{10\left(10^{2001}+1\right)}{10^{2002}+1}=\frac{10^{2002}+10}{10^{2002}+1}=\frac{10^{2002}+1+9}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)
\(B=\frac{10^{2002}+1}{10^{2003}+1}\Rightarrow10B=\frac{10\left(10^{2002}+1\right)}{10^{2003}+1}=\frac{10^{2003}+10}{10^{2003}+1}=\frac{10^{2003}+1+9}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)
Vì \(\frac{9}{10^{2002}+1}>\frac{9}{2^{2003}+1}\Rightarrow1+\frac{9}{10^{2002}+1}>1+\frac{9}{2^{2003}+1}\Rightarrow10A>10B\Rightarrow A>B\)
Vậy A > B