1+1:2+11-10+1
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Những câu hỏi liên quan
DN
1
4 tháng 7 2021
Giải:
a) Gọi dãy đó là A, ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\)
\(2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\)
\(2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^{2014}}\)
Vì \(\dfrac{1}{2}< 1;\dfrac{1}{2^{2014}}< 1\) nên \(\dfrac{1}{2}-\dfrac{1}{2^{2014}}< 1\)
\(\Rightarrow A< 1\)
b) \(A=\dfrac{10^{11}-1}{10^{12}-1}\) và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
Ta có:
\(A=\dfrac{10^{11}-1}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-10}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-1+9}{10^{12}-1}\)
\(10A=1+\dfrac{9}{10^{12}-1}\)
Tương tự:
\(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+1+9}{10^{11}+1}\)
\(10B=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\) nên \(10A< 10B\)
\(\Rightarrow A< B\)
A
2
13 tháng 5 2018
So sánh: mk làm luôn nè:
Ta có: \(\frac{10}{11}>\frac{10}{11+12};\frac{11}{12}>\frac{11}{11+12}\)
\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10}{11+12}+\frac{11}{11+12}\)
\(\Rightarrow\frac{10}{11}+\frac{11}{12}>\frac{10+11}{11+12}\)
MK KO BIẾT ĐÚNG KO NỮA NÊN BN CÓ THỂ THAM KHẢO CỦA CÁC BẠN KHÁC NHÉ.!!
CHÚC BẠN HỌC TỐT. ^_^
HA
1
13 tháng 5 2021
a) Ta có
A = n / n+1 = 1-(1/n+1)
A = n+2 / n+3 = 1-(1/n+3)
Vì 1/n+1 > 1/n+3
=> n/n+1 < n+2/n+3
=> A<B
21 tháng 5 2021
Câu 1:
\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\)
Câu 2:
\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\)
Câu 3:
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\)
Câu 4:
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)
\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=\dfrac{2}{7}.1\)
\(=\dfrac{2}{7}\)
S
21 tháng 5 2021
Câu 1
\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)
Câu 2
-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)
Câu 3
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)
Câu 4
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)=\(\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)=\(\dfrac{2}{7}.1\)=\(\dfrac{2}{7}\)
5 tháng 4 2023
\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)
có : `1-1/(11^99)<1`
\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)
hay `B<1/10`
3
3,5