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\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}+\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)+\left(\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{9}}\right)+...+\left(\frac{1}{\sqrt{82}}+...+\frac{1}{\sqrt{100}}\right)\)
\(>\frac{1}{\sqrt{1}}+\left(\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}\right)+\left(\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{9}}\right)+...+\left(\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\right)\)
\(>\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{10}{10}=10\)
\(\Leftrightarrow2-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{11}}\right)>0\)
Ta có: \(\frac{1}{2^{12}}-1=\left(\frac{1}{2}-1\right)\left(\frac{1}{2^{11}}+\frac{1}{2^{10}}+\frac{1}{2^9}+...+\frac{1}{2}+1\right)\)
\(\Rightarrow1+\frac{1}{2}+...+\frac{1}{2^{11}}=2\left(1-\frac{1}{2^{12}}\right)=2-\frac{1}{2^{11}}\)
\(\Rightarrow2-\left(1+\frac{1}{2}+...+\frac{1}{2^{11}}\right)=2-\left(2-\frac{1}{2^{11}}\right)=\frac{1}{2^{11}}>0\left(đpcm\right)\)
1-1/2-1/2^2-......-1/2^11
ta có:1-1/2-1/2^2-.....-1/2^11=1-(1/2+1/2^2+....+1/2^11)
A=1/2+1/2^2+1/2^3+...+1/2^11
2A=2.(1/2+1/2^2+1/2^3+...+1/2^11)
2A=2.1/2+2.1/2^2+....+2.1/2^11
2A-A=(1+1/2^2+1/2^3+...+1/2^10)-(1/2+1/2^2+1/2^3+....+1/2^11)
A=1-1/2^11=2048/2048-1/2048=2047/2048
vì 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-A
=> 1-(1/2+1/2^2+1/2^3+...+1/2^11)=1-2047/2048=2048/2048-2047/2048=1/2048=1/2^11
vậy 1-1/2-1/2^2-1/2^3-...-1/2^11=1/2^11
a,Cho x>y>0 chứng minh rằng x^2>y^2
b, Chứng minh rằng: Nếu lal<1;lb-1l<10 và la-cl<10 thì lab-cl<20
Ta có :
\(\dfrac{1}{\sqrt{1}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{2}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
.........................................
\(\dfrac{1}{\sqrt{99}}>\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\dfrac{1}{\sqrt{100}}=\dfrac{1}{\sqrt{100}}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+..........+\dfrac{1}{\sqrt{100}}>\dfrac{1}{10}+\dfrac{1}{10}+....+\dfrac{1}{10}=\dfrac{1}{10}.100=10\left(đpcm\right)\)
Ta có:
1/√1 > 1/10
1/√2 > 1/10
1/√3 > 1/10
....................
1/√99 > 1/10
1/√100 = 1/10
Cộng từng vế ta có:
1/√1 + 1/√2 + 1/√3 + ... + 1/√100 >100.1/0 = 10 (Đpcm)