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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
k mk nha
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)
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
=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)
=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)
\(A=\frac{10^8+2}{10^8-1}=-2\)
\(B=\frac{10^8-3}{10^8}=-3\)
\(\Rightarrow A>B\left(-2>-3\right)\)
\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
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