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A=\(\frac{98^{99}+1}{98^{89}+1}>1\) =>\(A=\frac{98^{99}+1}{98^{89}+1}>\frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}\)
\(=\frac{98.\left(98^{98}+1\right)}{98.\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C>D
\(A=\frac{3^{100}+1}{3^{99}+1}=\frac{\left(3^{99}+1\right)\times3-2}{3^{99}+1}=3-\frac{2}{3^{99}+1}\)
\(B=\frac{3^{99}+1}{3^{98}+1}=\frac{\left(3^{98}+1\right)\times3-2}{3^{98}+1}=3-\frac{2}{3^{98}+1}\)
Do 398 + 1 < 399 + 1
=> \(\frac{2}{3^{98}+1}>\frac{2}{3^{99}+1}\)
=> A > B
Ta co : 72/73 = 1 - 1/73 ; 98/99 = 1 - 1/99
Vì 1/73 > 1/99 suy ra 1 - 1/73 < 1 - 1/99 hay 72/73 < 98/99
Ta có: \(\frac{72}{73}=1-\frac{1}{73}\);\(\frac{98}{99}=1-\frac{1}{99}\)
Vì \(\frac{1}{73}>\frac{1}{99}\Rightarrow1-\frac{1}{73}< 1-\frac{1}{99}\)
\(\Rightarrow\frac{72}{73}< \frac{98}{99}\)
Vậy \(\frac{72}{73}< \frac{98}{99}\)
Vì 45<47 nên \(\frac{45}{47}< 1\)
Ta có:\(\frac{45}{47}< \frac{\text{45+1}}{47+1}=\frac{46}{48}\)
Vậy \(\frac{45}{47}< \frac{46}{48}\).
(k mik nha bn!)
Cái này mìk làm theo quán tính của mìk thui nha!
Ta thấy:
* \(45< 46\)
* \(47< 48\)
\(\Rightarrow\frac{45}{47}< \frac{46}{48}\)
Đó là theo cách mìk nghĩ thui, k biết có đúng ko nữa...........?
Bài 1:
Ta thấy A < 1
=> A = \(\frac{17^{18}+1}{17^{19}+1}< \frac{17^{18}+1+16}{17^{19}+1+16}=\frac{17^{18}+17}{17^{19}+17}=\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\frac{17^{17}+1}{17^{18}+1}=B\)
Vậy A < B
Bài 2:
Ta thấy C < 1
=> C = \(\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=D\)
Vậy C < D
a) \(\frac{8}{9}=1-\frac{1}{9}\)
\(\frac{108}{109}=1-\frac{1}{109}\)
Vì \(\frac{1}{9}>\frac{1}{109}\)
Nên \(1-\frac{1}{9}< 1-\frac{1}{109}\)
Vậy \(\frac{8}{9}< \frac{108}{109}\)
b)
\(\frac{97}{100}=\frac{97\cdot99}{100\cdot99}\)
\(\frac{98}{99}=\frac{98\cdot100}{99\cdot100}\)
\(\Rightarrow\frac{97}{100}< \frac{98}{99}\)
\(A=\frac{-\left(98^{98}+1\right)}{-\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}\)
\(B=\frac{98^{99}+1}{98^{89}+1}\)
A-1=\(\frac{98^{98}-98^{88}}{98^{88}+1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}\)
B-1=\(\frac{98^{99}-98^{89}}{98^{89}+1}=\frac{98^{89}.\left(98^{10}-1\right)}{98^{89}+1}\)
=>\(\frac{A-1}{B-1}=\frac{98^{88}.\left(98^{10}-1\right)}{98^{88}+1}.\frac{98^{89}+1}{98^{89}.\left(98^{10}-1\right)}=\frac{98^{89}+1}{98.\left(98^{88}+1\right)}=\frac{98^{89}+1}{98^{89}+98}< 1\)
->A-1<B-1
->A<B
So sánh bằng nhân chéo :
Ta có : 99 . 47 = 4653
98 . 48 = 4704
So sánh kết quả : 4653 < 4704 . Nên :
\(\frac{99}{98}\) < \(\frac{48}{47}\)
Kết luận : Vậy \(\frac{99}{98}\) < \(\frac{48}{47}\)
\(\frac{99}{98}\) = \(\frac{4653}{4606}\) ; \(\frac{48}{47}\) = \(\frac{4704}{4606}\)
So sánh \(\frac{4653}{4606}\) < \(\frac{4704}{4606}\) vậy \(\) \(\frac{99}{98}\) < \(\frac{48}{47}\) .