Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
c , Ta có : 201/195 - 1 = 6/195
203/197 - 1 = 6/197
Vì 6/195 > 6/ 197 nên 201/195>203/197
d , Ta có : 1998/1995 - 1 = 3/1995
2005/2002 - 1 = 3/2002
Vì 3/1995 > 3/2002 nên 1998/1995 > 2005/2002
c , Ta có : 201/195 - 1 = 6/195
203/197 - 1 = 6/197
Vì 6/195 > 6/ 197 nên 201/195>203/197
d , Ta có : 1998/1995 - 1 = 3/1995
2005/2002 - 1 = 3/2002
Vì 3/1995 > 3/2002 nên 1998/1995 > 2005/2002
Đúng 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
a= 202 x 204 = 202 x (203+1)=202 x 203 + 202
b=203 x 203 = (202+1) x 203 = 202 x 203 + 203
Vì 203>202 => 202x 203 + 202<202x203 +203
=>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
=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)
=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)
so sánh bằng phân số trung gian
vì \(\frac{106}{204}\)> \(\frac{106}{206}\)nên \(\frac{106}{204}\)> \(\frac{107}{206}\)
Ta có: \(\frac{2}{3}=1-\frac{1}{3}\)
\(\frac{203}{204}=1-\frac{1}{204}\)
Vì \(\frac{1}{3}>\frac{1}{204}\) nên \(1-\frac{1}{3}< 1-\frac{1}{204}\)
hay \(\frac{2}{3}< \frac{203}{204}\)
Ta có :
\(\frac{2}{3}=1-\frac{1}{3}\)
\(\frac{203}{204}=1-\frac{1}{204}\)
Ta thấy :
\(\frac{1}{3}>\frac{1}{204}\left(3< 204\right)\)
\(\Rightarrow1-\frac{1}{3}< 1-\frac{1}{204}\)
\(\Rightarrow\frac{2}{3}< \frac{203}{204}\)
~ Thiên Mã ~