Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Do A = 98 99 + 1 98 89 + 1 > 1 nên A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 98 98 + 1 98 98 88 + 1 = 98 98 + 1 98 88 + 1 = B
Vậy A > B
a) Do A = 98 99 + 1 98 89 + 1 > 1 nên
A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 ( 98 98 + 1 ) 98 ( 98 88 + 1 ) = 98 98 + 1 98 88 + 1 = B
Vậy A > B
b) Do C = 100 2008 + 1 100 2018 + 1 < 1 nên
C= 100 2008 + 1 100 2018 + 1 > 100 2008 + 1 + 99 100 2018 + 1 + 99 = 100 ( 100 2007 + 1 ) 100 ( 100 2017 + 1 ) = 100 2007 + 1 100 2017 + 1 = D
Vậy C > D.
Sửa đề: \(C=\dfrac{17^{99}+1}{17^{99}-1}\)
\(C=\dfrac{17^{99}-1+2}{17^{99}-1}=1+\dfrac{2}{17^{99}-1}\)
\(D=\dfrac{17^{98}-1+2}{17^{98}-1}=1+\dfrac{2}{17^{98}-1}\)
17^99>17^98
=>17^99-1>17^98-1
=>C<D
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
Bài 6.7*
Ta có : \(\dfrac{17^{18}+1}{17^{19}+1}< 1\)
\(\Rightarrow A=\dfrac{17^{18}+1}{17^{19}+1}< \dfrac{17^{18}+1+16}{17^{19}+1+16}=\dfrac{17^{18}+17}{17^{19}+17}=\dfrac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}=\dfrac{17^{17}+1}{17^{18}+1}=B\)
\(\)Vậy A < B
Bài 6.6*
Ta có : \(\dfrac{98^{99}+1}{98^{89}+1}>1\)
\(\Rightarrow C=\dfrac{98^{99}+1}{98^{89}+1}>\dfrac{98^{99}+1+97}{98^{89}+1+97}=\dfrac{98^{99}+98}{98^{89}+98}=\dfrac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\dfrac{98^{98}+1}{98^{88}+1}=D\)
Vậy C > D