Chứng tỏ rằng:A=\(\frac{10}{27}\)+\(\frac{9}{16}\) +\(\frac{11}{34}\) <2
Chứng tỏ :
S = \(\frac{1}{201}\)+ \(\frac{1}{202}\)+.........+\(\frac{1}{399}\)+\(\frac{1}{400}\)>\(\frac{1}{2}\)
A = \(\frac{10}{27}\)+ \(\frac{9}{16}\)+ \(\frac{11}{34}\)< 2
Cho A=\(\frac{10}{17}+\frac{8}{15}+\frac{11}{16}\).Chứng tỏ rằng A< 2
Cho A = 10/27 + 9/16 + 11 / 34 . Chứng tỏ rằng a < 2
ta có: \(\frac{10}{27}< \frac{10}{30}=\frac{1}{3}\)
\(\frac{9}{16}< \frac{9}{9}=1\)
\(\frac{11}{34}< \frac{11}{22}=\frac{1}{2}\)
=>A<\(\frac{1}{3}+1+\frac{1}{2}\)<2
vậy A<2
Tính \(A=16-\frac{\frac{-2}{9}+\frac{-2}{10}+\frac{-2}{11}+...+\frac{-2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(A=16-\frac{\left(-2\right)\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\cdot\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)
\(A=16-\frac{-2}{\frac{1}{3}}=16-\left(-6\right)=22\)
Vậy A = 22
Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(A=\frac{1}{10}+\frac{99}{100}=1\)
=> A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(A=\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+... +\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\Rightarrow A>1\)
Ta thấy:1/10;1/11;1/12;1/13;...;1/99>1/100
=)1/10+1/11+1/12+1/13+...+1/100>1/100+1/100+1/100+1/100..+1/100=1/100.100=1
Vậy A>1
Cho tổng A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...\frac{1}{99}+\frac{1}{100}\)
Chứng tỏ rằng A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)
\(=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(=\frac{1}{10}+\frac{90}{100}>1\)
\(A>1\left(đpcm\right)\)
Tính giá trị biểu thức A biết:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
Ta có:
\(A=16-\frac{-\frac{2}{9}-\frac{2}{10}-\frac{2}{11}-...-\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(\Rightarrow A=16+\frac{\frac{2}{9}+\frac{2}{10}+\frac{2}{11}+...+\frac{2}{2020}}{\frac{1}{27}+\frac{1}{30}+\frac{1}{33}+...+\frac{1}{6060}}\)
\(\Rightarrow A=16+\frac{2\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}{\frac{1}{3}\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{2020}\right)}\)
\(\Rightarrow A=16+\frac{2}{\frac{1}{3}}\)
\(\Rightarrow A=16+\left(2:\frac{1}{3}\right)\)
\(\Rightarrow A=16+\left(2.3\right)\)
\(\Rightarrow A=16+6\)
\(\Rightarrow A=22\)
Vậy\(A=22\)
A = 16 + (2/9+2/10+....+2/2020)/(1/27+1/30+.....+1/6060)
= 16 + 6
= 22
Tk mk nha
cho a=\(\frac{9\frac{3}{4}:5,2+3,4.2\frac{7}{34}:1\frac{9}{16}}{0,31.8\frac{2}{5}-5,61:27\frac{1}{2}}\)
a, rút gọn a
b, tìm 2,5% của a
