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Tham khảo:Câu hỏi của Trần Trí Trung - Toán lớp 6 - Học toán với OnlineMath
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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
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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
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\(A=\frac{10^{2001}+1}{10^{2002}+1}=\frac{\left(10^{2001}+1\right)\left(10^{2003}+1\right)}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}=\frac{10^{4004}+10^{2001}+10^{2003}+1}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}\)
\(B=\frac{10^{2002}+1}{10^{2003}+1}=\frac{\left(10^{2002}+1\right)\left(10^{2002}+1\right)}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}=\frac{10^{4004}+2.10^{2002}+1}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}\)
Vì 102001 + 102003 < 2.102002 nên A < B
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\(A=\dfrac{10^{2004}+1}{10^{2003}+1}>\dfrac{10^{2004}+1+9}{10^{2003}+1+9}=\dfrac{10^{2004}+10}{10^{2003}+10}.\\ =\dfrac{10\left(10^{2003}+1\right)}{10\left(10^{2002}+1\right)}=\dfrac{10^{2003}+1}{10^{2002}+1}=B.\\ \Rightarrow A>B.\)