CHỨNG MINH RẰNG: + ++.....+ >10
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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
\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}+\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}\right)+\left(\frac{1}{\sqrt{5}}+...+\frac{1}{\sqrt{9}}\right)+...+\left(\frac{1}{\sqrt{82}}+...+\frac{1}{\sqrt{100}}\right)\)
\(>\frac{1}{\sqrt{1}}+\left(\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}\right)+\left(\frac{1}{\sqrt{9}}+...+\frac{1}{\sqrt{9}}\right)+...+\left(\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\right)\)
\(>\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{10}{10}=10\)
Ta có:\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\)> \(\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)[ 90 p/s \(\frac{1}{100}\)]
= \(\frac{1}{10}+\frac{90}{100}=\frac{10}{100}+\frac{90}{100}\)=\(\frac{100}{100}=1\)
Vậy \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\)>1
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\)\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\)
A = y2 -5y + 10 = y2 - 2*y*\(\frac{5}{2}\)+ \(\frac{25}{4}\)+ \(\frac{15}{4}\)= (y - \(\frac{5}{2}\))2 + \(\frac{15}{4}\)
vi (y - \(\frac{5}{2}\))2 >= 0
nên (y - \(\frac{5}{2}\))2 + \(\frac{15}{4}\) > 0
A=y^2-5y+10
=> y^2-2.y.\(\frac{5}{2}\)+\(\frac{25}{4}-\frac{25}{4}+10\)
=>\(\left(y-\frac{5}{2}\right)^2-\frac{25}{4}+10\)
=>\(\left(y-\frac{5}{2}\right)^2+\frac{15}{4}\)
vì \(\left(y-\frac{5}{2}\right)^2\)>0 \(\forall\)y
=>\(\left(y-\frac{5}{2}\right)^2+\frac{15}{4}\)\(\ge\)\(\frac{15}{4}\)
Chứng minh rằng: mày bị ngáo
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