B1:
Cho A = \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Hãy chứng tỏ rằng A > 1
B2:
So sánh: \(B=\) \(\frac{20^{10}+1}{20^{10}-1}\)và \(C=\frac{20^{10}-1}{20^{10}-3}\)
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So sánh A = \(\frac{10^{19}+1}{10^{20}+1}\) và B = \(\frac{10^{20}+1}{10^{21}+1}\) ?
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
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so sánh
A=\(\frac{10^{19}+1}{10^{20}+1}\)và B=\(\frac{10^{20}+1}{10^{21}+1}\)
\(B=\frac{10^{20}+1}{10^{21}+1}< 1\)
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
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So sánh A và B:
\(A=\frac{20^{10}+1}{20^{10}-1};B=\frac{20^{10}-1}{20^{10}-3}\)
So sánh A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
ta thấy B>1 nên B=\(\frac{20^{10}-1}{20^{10}-3}\)>\(\frac{20^{10}-1+2}{20^{100}-3+2}\)=\(\frac{20^{10}+1}{20^{10}-1}\)=A
vậy B>A
nếu ko hiểu thì tham khảo trong SBT lớp 6 bài so sánh PS ấy
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So sánh : A=\(\frac{20^{10}+1}{20^{10}-1}\)và B=\(\frac{20^{10}-1}{20^{10}-3}\)
Vì \(20^{10}-1>20^{10}-3\)
\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}>\frac{20^{10}-1+2}{20^{10}-3+2}=\frac{20^{10}+1}{20^{10}-1}=A\)
vậy \(A< B\)
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So sánh phân số sau : A=\(\frac{^{10^{19}}+1}{10^{20}+1}\)và B = \(\frac{10^{20}+1}{10^{21}+1}\)
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
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So sánh A = \(\frac{10^{19}+1}{10^{20}+1}\) và B = \(\frac{10^{20}+1}{10^{21}+1}\) ?
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
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So sánh A và B biết
A=\(\frac{10^{19}+1}{10^{20}+1}\)
B=\(\frac{10^{20}+1}{10^{21}+1}\)
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
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So sánh : \(A =\frac{20^{10} +1}{20^{10}-1} ; B =\frac{20^{10} -1}{20^{10} -3}\)
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> A < B
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