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10A=\(\frac{10^{20}+10}{10^{20}+1}\)=\(\frac{10^{20}+1+9}{10^{20}+1}\)=\(1\)+\(\frac{9}{10^{20}+1}\)
10B=\(\frac{10^{21}+10}{10^{21}+1}\)=\(\frac{10^{21}+1+9}{10^{21}+1}\)=\(1\)+\(\frac{9}{10^{21}+1}\)
Vì \(\frac{9}{10^{20}+1}\)>\(\frac{9}{10^{21}+1}\)nên 10A>10B\(\Rightarrow\)A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(B=\frac{10^{20}+1}{10^{21}+1}\)<1
\(\Rightarrow B=\frac{10^{20}+1}{10^{21}+1}\)<\(\frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
\(\Rightarrow\)B<A hay A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(\frac{10^{20}+1}{10^{21}+1}\)
=> 10A = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Đặt B = \(\frac{10^{21}+1}{10^{22}+1}\)
=> 10B = \(\frac{10^{22}+10}{10^{22}+1}=1+\frac{9}{10^{22}+1}\)
Vì \(\frac{9}{10^{21}+1}>\frac{9}{10^{22}+1}\)
=> \(1+\frac{9}{10^{21}+1}>1+\frac{9}{10^{22}+1}\)
=> 10A > 10B
=> A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta chứng minh bài toán phụ:
Với a<b thì\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(c\inℕ^∗\right)\)
Ta có: \(a< b\)
\(\Rightarrow ac< bc\)
\(\Rightarrow ac+ba< bc+ba\)
\(a\left(b+c\right)< b.\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)
đpcm
Áp dụng vào bài toán ta có:
\(\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}=\frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}\)
Vậy \(\frac{10^{19}+1}{10^{20}+1}>\frac{10^{20}+1}{10^{21}+1}\)
Tham khảo nhé~
Đặt \(A=\frac{10^{19}+1}{10^{20}+1}\)
\(\Rightarrow10A=\frac{10^{20}+10}{10^{20}+1}=\frac{10^{20}+1+9}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
\(B=\frac{10^{20}+1}{10^{21}+1}\)
\(\Rightarrow10B=\frac{10^{21}+10}{10^{21}+1}=\frac{10^{21}+1+9}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
\(\Rightarrow\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\)
\(\Rightarrow1+\frac{9}{10^{20}+1}>1+\frac{9}{10^{21}+1}\)
\(\Rightarrow10A>10B\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 3)(y - 3) = 9 = 1.9 = 3.3
Lập bảng:
x - 3 | 1 | -1 | 3 | -3 | 9 | -9 |
y - 3 | 9 | -9 | 3 | -3 | 1 | -1 |
x | 4 | 2 | 6 | 0 | 12 | -3 |
y | 12 | -6 | 6 | 0 | 4 | 2 |
Vậy ...
b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
Vậy A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : B = \(\frac{9^{19}+1}{9^{20}+1}\)< \(\frac{9^{19}+1+8}{9^{20}+1+8}\)= \(\frac{9^{19}+9}{9^{20}+9}\)= \(\frac{9\left(9^{18}+1\right)}{9\left(9^{19}+1\right)}\)= \(\frac{9^{18}+1}{9^{19}+1}\)= A
Vậy A > B
b) Ta có : B = \(\frac{10^{2018}-1}{10^{2019}-1}\)> \(\frac{10^{2018}-1-9}{10^{2019}-1-9}\)= \(\frac{10^{2018}-10}{10^{2019}-10}\)= \(\frac{10\left(10^{2017}-1\right)}{10\left(10^{2018}-1\right)}\)= \(\frac{10^{2017}-1}{10^{2018}-1}\)= A
Vậy A < B.
NHỚ K CHO MK VỚI NHÉ !!!!!!!!
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Chứng minh nếu a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Do a/b < 1 => a < b
=> am < bm
=> am + ab < bm + ab
=> a.(b+m) < b.(a+m)
=> a/b < a+m/b+m
Áp dụng điều trên ta có: B = 1020 + 1/ 1021 + 1 < 1
=> B < 1020 + 1 + 9/1021 + 1 + 9
=> B < 1020 + 10/1021 + 10
=> B < 10.(1019 + 1)/10.(1020 + 1)
=> B < 1019+1/1020+1 = A
=> B < A
b) n + 1 chia hết cho n - 2
=> n - 2 + 3 chia hết cho n - 2
Do n - 2 chia hết cho n - 2
=> 3 chia hết cho n - 2
=> n - 2 thuộc { 1 ; -1 ; 3 ; -3}
=> n thuộc { 3 ; 1 ; 5 ; -1}
Vậy n thuộc { 3 ; 1 ; 5 ; -1}
Ta thấy B < 1 và 9 > 1 nên ta có:
B < 1020 + 1 + 9 / 1021 + 1 + 9
=> B < 1020 + 10 / 1021 + 10
=> B < 10(1019 + 1) / 10(1020 + 1)
=> B < 1019 + 1 / 1020 + 1 = A
=> B < A