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\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)(1)
Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)
\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{10}}\)
Thay A vào (1)
\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)
\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
Ta có: 210 < 211
\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)
A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102
=1+0+0+....+102=103
b) |1-2x|>7
=> 1-2x>7 hoặc 1-2x<-7
=> 2x<-6 hoặc 2x>8
=> x<-3 hoặc x>4
1>1/1*2
1/22>1/2*3
1/32>1/3/4
.....................
1/1002>1/100*101
=>1-1/22-...-1/1002>1/1*2-1/2*3-.....-1/100*101=1-1/2-1/2+1/3-1/3+......-1/100+1/101=1/101
vậy 1-1/22-....-1002
study well
k nha
ai k đúng cho mk thì mk trả lại gấp đôi và ngược lại
ai ghé qua nhớ để lại 1 k đúng
ủng hộ mk nha
a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm
b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)
Vì \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{100^2}<\frac{1}{99.100}\)nên \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}<\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\)
=> \(B<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)
=> 1 - B > \(1-\left(1-\frac{1}{100}\right)=\frac{1}{100}\) => đpcm
Cho P=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}\). Chứng tỏ rằng \(\frac{1}{15}< P< \frac{1}{10}\)
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
______________________________________________
\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
Ta có : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)
= \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Thấy : \(\frac{1}{11}>\frac{1}{100}\)
\(\frac{1}{12}>\frac{1}{100}\)
...
\(\frac{1}{99}>\frac{1}{100}\)
Cộng từng vế : \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+...+\frac{1}{100}\)( 90 SH 1/100)
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{9}{10}\)
=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}\)
=> Tổng trên > 1