\(\dfrac{98}{56}\) + \(\dfrac{99}{88}\)
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LL
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)
3 tháng 3 2023
a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)
\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)
98^88+1>98^99+1
=>A<B
b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)
\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)
2022^2023>2022^2021
=>2022^2023+2022^2>2022^2021+2022^2
=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)
=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)
=>C>D
MM
2
3 tháng 3 2022
Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
SG
3
17 tháng 3 2018
Vì C= \(\dfrac{98^{99}+1}{98^{89}+1}\)>1 thì nên áp dụng tính chất . Nên \(\dfrac{a}{b}\)>1 thì \(\dfrac{a}{b}\)>\(\dfrac{a+m}{b+m}\) ( a∈ N , b và m ∈ N✳) Ta có : C= \(\dfrac{98^{99}+1}{98^{89}+1}\)> \(\dfrac{98^{99}+1+97}{98^{89}+1+97}\)= \(\dfrac{98^{99}+98}{98^{89}+98}\) = \(\dfrac{98.98^{98}+98.1}{98.98^{88}+98.1}\) = \(\dfrac{98.\left(98^{98}+1\right)}{98.\left(98^{88}+1\right)}\)= \(\dfrac{98^{98}+1}{98^{88}+1}\)= B ⇔ Vậy \(\dfrac{98^{99}+1}{98^{89}+1}\)< \(\dfrac{98^{89}+1}{98^{88}+1}\) nên C<D
HD
0
17 tháng 9 2017
Áp dụng tính chất \(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có :
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{98}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
HD
17 tháng 9 2017
Ta áp dụng tính chất :
\(\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ta có:
\(B=\dfrac{10^{99}+1}{10^{89}+1}>\dfrac{10^{99}+1+9}{10^{89}+1+9}=\dfrac{10^{99}+10}{10^{89}+10}=\dfrac{10\left(10^{98}+1\right)}{10\left(10^{88}+1\right)}=\dfrac{10^{98}+1}{10^{88}+1}=A\)
\(\Leftrightarrow B>A\)
Chúc bạn học tốt!
4 tháng 4 2017
Đặt : \(B=\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
\(B=\left(\dfrac{99}{1}+1\right)+\left(\dfrac{98}{2}+1\right)+...+\left(\dfrac{1}{99}+1\right)-99\)
\(B=\dfrac{100}{1}+\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}-99\)
\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\left(100-99\right)\)
\(B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+\dfrac{100}{100}\)
\(B=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)
Ta có : \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}{100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)}=\dfrac{1}{100}\)
\(8624/4928+5544/4928=14168/4928=23/8\) nhé =)