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đặt \(A=\frac{10^{18}+1}{10^{19}+1};B=\frac{10^{19}+1}{10^{20}+1}\)
ta có: \(10A=\frac{10^{19}+1+9}{10^{19}+1}=1+\frac{9}{10^{19}+1}\)
\(10B=\frac{10^{20}+1+9}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
mà \(\frac{9}{10^{19}+1}>\frac{9}{10^{20}+1}\)
=> 10A >10B
=> A > B
Bài 1:
a) \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)
\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)
\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)
\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)
\(x-\frac{8}{11}=\frac{3}{11}\)
\(x=1\)
b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)
\(\frac{1}{x+1}=\frac{1}{18}\)
=> x + 1 =18
x = 17
bài 2 ko bk lm, xl nha
a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)
\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)
Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)
b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)
\(\frac{2021}{2020}=1+\frac{1}{2020}\)
Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)
c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)
\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)
Đến đây tự so sánh rồi nhé
Ta luôn có nếu a>0; b>0 thì \(\frac{a}{b}< \frac{a+m}{b+m}\left(m\in N\right)\)
Áp dụng vào bài toán ta thấy 1011-1 > 0 và 1012-1 > 0 nên
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10.\left(10^{10}+1\right)}{10.\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
Vậy A < B
Xin lỗi bn nhé bài toán phụ phía trên đang còn 1 đk nữa là a<b
\(B< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
Vậy A > B
Áp dụng bất đẳng thức :
\(\frac{a}{b}< 1\Leftrightarrow\frac{a}{b}< \frac{a+m}{b+m}\)
Ta có :
\(B=\frac{10^{2012}+1}{10^{2013}+1}< \frac{10^{2012}+1+9}{10^{2013}+1+9}=\frac{10^{2012}+10}{10^{2013}+10}=\frac{10\left(10^{2011}+1\right)}{10\left(10^{2012}+1\right)}=\frac{10^{2011}+1}{10^{2012}+1}=A\)
\(\Leftrightarrow B< A\)
Nhân cả hai tử của \(A\)và \(B\)với 2 , ta được :
\(10A=10.\left(\frac{10^{2016}+1}{10^{2017}+1}\right)=\frac{10^{2017}+1+9}{10^{2017}+1}=1+\frac{9}{2^{2017}+1}\)
\(10B=10\left(\frac{10^{2017}+1}{10^{2018}+1}\right)=\frac{10^{2018}+10}{10^{2018}+1}=\frac{10^{2018}+1+9}{10^{2018}}=1+\frac{9}{10^{2018}+1}\)
Vì \(1=1;9=9\)
\(\Rightarrow\)Ta so sánh mẫu , ta có:
\(10^{2017}< 10^{2018}\)
\(\Rightarrow10^{2017}+1< 10^{2018}+1\)
\(\Rightarrow1+\frac{9}{10^{2017}+1}>1+\frac{9}{10^{2018}+1}\)
\(\Rightarrow10A>10B\)
Hay \(A>B\)
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
NÊN \(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
VẬY B<A