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

đúng rồi

\(A=\frac{10^{2001}+1}{10^{2002}+1}=\frac{\left(10^{2001}+1\right)\left(10^{2003}+1\right)}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}=\frac{10^{4004}+10^{2001}+10^{2003}+1}{\left(10^{2002}+1\right)\left(10^{2003}+1\right)}\)

\(B=\frac{10^{2002}+1}{10^{2003}+1}=\frac{\left(10^{2002}+1\right)\left(10^{2002}+1\right)}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}=\frac{10^{4004}+2.10^{2002}+1}{\left(10^{2003}+1\right)\left(10^{2002}+1\right)}\)

Vì 10²⁰⁰¹ + 10²⁰⁰³ < 2.10²⁰⁰² nên A < B

Không nhầm là quy đồng phân số A nhân với 10

Đặt A = \(\frac{10^{2001}+1}{10^{2002}+1}\) ; B = \(\frac{10^{2002}+1}{10^{2003}+1}\)

10A = \(\frac{10\left(10^{2001}+1\right)}{10^{2002}+1}\) \(=\frac{10^{2002}+1+9}{10^{2002}+1}\)

10B = \(y=\frac{10\left(10^{2002}+1\right)}{10^{2013}+1}\)\(=\frac{10^{2003}+1+9}{10^{2003}+1}\)

Vì \(y=\frac{10^{2002}+1+9}{10^{2002}+1}\) > \(\frac{10^{2003}+1+9}{10^{2003}+1}\) nên 10A > 10B nên A > B

Cũng ko chắc nữa -_-

B lon hon nha

Ta có: A<B là vì khi tính ra thì được A âm và B dương

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

\(A=\frac{10^{2001}+1}{10^{2002}+1}\Rightarrow10A=\frac{10.\left(10^{2001}+1\right)}{10^{2002}+1}=\frac{10^{2002}+10}{10^{2002}+1}\)

\(10A=\frac{10^{2002}+1+9}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)

\(B=\frac{10^{2002}+1}{10^{2003}+1}\Rightarrow10B=\frac{10.\left(10^{2002}+1\right)}{10^{2003}+1}=\frac{10^{2003}+10}{10^{2003}+1}\)

\(10B=\frac{10^{2003}+1+9}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)

Vì \(10^{2002}+1<10^{2003}+1\Rightarrow\frac{9}{10^{2002}+1}>\frac{9}{10^{2003}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Một bạn A > B, một bạn B < A mình biết chọn cái nào đây

a<b

k minh

Tham khảo:Câu hỏi của Trần Trí Trung - Toán lớp 6 - Học toán với OnlineMath

ghi lộn phải bé chứ không phải lớn

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