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Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)
Ta có:
\(\left(\frac{-1}{8}\right)^{100}=\frac{\left(-1\right)^{100}}{8^{100}}=\frac{1}{\left(2^3\right)^{100}}=\frac{1}{2^{300}}\)
\(\left(\frac{-1}{4}\right)^{200}=\frac{\left(-1\right)^{200}}{4^{200}}=\frac{1}{\left(2^2\right)^{100}}=\frac{1}{2^{200}}\)
Vì \(2^{300}>2^{200}\)\(\Rightarrow\frac{1}{2^{300}}< \frac{1}{2^{200}}\)
\(\Rightarrow\left(\frac{-1}{8}\right)^{^{100}}< \left(\frac{-1}{4}\right)^{200}\)
ta có:A=\(\frac{-3}{2^2}.\frac{-8}{3^2}....\frac{-9999}{100^2}\)
A có 99 thừa số âm
=>A<0
\(=>-A=\frac{3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100.100}\)
=>\(-A=\frac{101}{100.2}=\frac{101}{200}>\frac{100}{200}=\frac{1}{2}=>-A>\frac{1}{2}=>A<-\frac{1}{2}\)
tick nhé
Bài 1: \(\left(\frac{-1}{16}\right)^{100}=\frac{1}{\left(2^4\right)^{100}}=\frac{1}{2^{400}}>\frac{1}{2^{500}}=\left(\frac{-1}{2}\right)^{500}.\)
Bài 2: \(100^{99}+1>100^{68}+1\Rightarrow\frac{1}{100^{99}+1}< \frac{1}{100^{68}+1}\Rightarrow\frac{-99}{100^{99}+1}>\frac{-99}{100^{68}+1}\)
\(\Rightarrow100+\frac{-99}{100^{99}+1}>100+\frac{-99}{100^{68}+1}\Rightarrow\frac{100^{100}+1}{100^{99}+1}>\frac{100^{69}+1}{100^{68}+1}\)
\(K=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}...\frac{-9999}{10000}=\left(-1\right)^{99}.\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}=-\frac{1}{100}.\frac{101}{2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)
ta có:1/8^100
-1/4^200=(-1/4^2)^100=1/16^100
=>1/8^100 >1/16^100
=>1/8^100 >-1/4^200