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

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

B lon hon nha

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

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

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