\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì 10⁸-1 > 10⁸-3

=>\(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

=>\(1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\Rightarrow A< B\)

Ta có: A = 10⁸ + 2/ 10⁸ - 1 = 3/10⁸ - 1                           

              B = 10⁸ / 10⁸ - 3 = 3 / 10⁸ -3             

Vì 3 / 10⁸ - 1 < 3 / 10⁸ -3 nên           

Nên A< B

\(A=\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}>1+\frac{3}{10^8-3}=\frac{10^8}{10^8-3}=B\)

vậy A>B

Ta có:\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}\)

\(\Rightarrow A=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}\)

\(\Rightarrow A=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}\)

\(\Rightarrow B=1+\frac{3}{10^8-3}\)

Vì \(\frac{3}{10^8-1}>\frac{3}{10^8-3}\Rightarrow A>B\)

Mình làm câu a) nha!!!

+) \(A=2009^{2010}+2009^{2009}\)

        \(=2009^{2009}.\left(2009+1\right)\)

        \(=2009^{2009}.2010\)

+) \(B=2010^{2010}=2010^{2009}.2010\)

Vì \(2010^{2009}>2009^{2009}\)nên \(2010^{2009}.2010>2009^{2009}.2010\)hay \(B>A\)

Vậy \(A< B\)

Hok tốt nha^^

A=10^8+2/10^8-1=10^8-1+3/10^8-1 
=10^8-1/10^8-1+3/10^8-1=1+3/10^8-1=1/3/... 
B=10^8/10^8-3=10^8-3+3/10^8-3 
=10^8-3/10^8-3+3/10^8-3=1+3/10^8-3=1/3/... 
tu (1) va (2) =>1/3/10^8-1<1/3/10^8-3(vi phân so nao 
co mau be hon thi phân so do lon hon nen 10^8-1>10^8-3) 
Hay A<B 

Nguồn: cho hỏi bài toán lớp 6? | Yahoo Hỏi & Đáp

A=10^8+2/10^8-1=10^8-1+3/10^8-1 
=10^8-1/10^8-1+3/10^8-1=1+3/10^8-1=1/3/... 
B=10^8/10^8-3=10^8-3+3/10^8-3 
=10^8-3/10^8-3+3/10^8-3=1+3/10^8-3=1/3/... 
tu (1) va (2) =>1/3/10^8-1<1/3/10^8-3(vi phân so nao 
co mau be hon thi phân so do lon hon nen 10^8-1>10^8-3) 
Hay A<B 
2) (1/26+1/50) : 1 + 1....(con khúc cuối do ban tu gjaj quyết di 
tinh xong la hết bai 2) 
giai thích bai 2) vi sao mjnh laj lam (1/26+1/50) : 1 + 1 vi 
mjnh lay so dau + so cuoi chia khoang cach so hang va + 1, 
do la cong thuc tinh, ban hay co gang nho va van dung vao thi cu , 
cho mjnh 5 * nhe :D)

 xin chao toi la học sinh lp 6 ten toi la Tran Thi Nhu Hoai toi dac biet hoc gioi mon anh van vi me toi la mot giao vien day anh toi khong thich mon toan lam vi no kha kho so voi thuc luc cua ban than toi nhung toi lai co ratnhieu vo sach mon toan mong se co nhieu bạn lien he voi toi de duoc nhan nhieu vo toan hon nua .Bai toan tren qua kho nen t khong the lam ngay duocj mong cac ban thong cam toi hua se giai bai toan nay sau khoang 50 phut toi

A=10^8+2/10^8-1

A=2/-1=-2

B=10^8/10^8-3

b=1/-3=-3

mà -2>-3    >>A>B

\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)

Nhận thầy 10⁸ - 1 > 10⁸ - 3

=> \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

=> \(1+\frac{3}{10^8-1}< \frac{3}{10^8-3}+1\)

=> A < B

b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)

17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)

Nhận thầy 17²⁰³ + 1 < 17²⁰⁴ + 1

=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)

=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)