Cho hai phân số:\(A=\frac{10^{19}+1}{10^{20}+1}\) và \(B=\frac{10^{20}+1}{10^{21}+1}\)
So sánh A và B (trình bày cả lời giải cho mình nhé!)
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Áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\) (a;b;c \(\in\) N*)
Ta có:
\(B=\frac{10^{20}+1}{10^{21}+1}< \frac{10^{20}+1+9}{10^{21}+1+9}=\frac{10^{20}+10}{10^{21}+10}\)
\(B< \frac{10.\left(10^{19}+1\right)}{10.\left(10^{20}+1\right)}=\frac{10^{19}+1}{10^{20}+1}=A\)
=> A > B
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
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
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
Ta thấy:A=\(\frac{10^{19}+1}{10^{20}+1}\)=>10A=\(\frac{10^{20}+10}{10^{20}+1}\)
=>10A=\(\frac{10^{20}+1+9}{10^{20}+1}\)
=>10A=1+\(\frac{9}{10^{20}+1}\)
Ta thấy:B=\(\frac{10^{20}+1}{10^{21}+1}\)
=>10B=\(\frac{10^{21}+10}{10^{21}+1}\)
=>10B=\(\frac{10^{21}+1+9}{10^{21}+1}\)
=>10B=1+\(\frac{9}{10^{21}+1}\)
Do \(\frac{9}{10^{20}+1}\)> \(\frac{9}{10^{21}+1}\)=>A > B
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}
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 !!!!!!!!
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
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