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: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

\(\frac{1}{2}>\frac{1}{3}\\ \Rightarrow\left(\frac{1}{2}\right)^{200}>\left(\frac{1}{3}\right)^{200}\)

Vì 200=200 và 1/2 >1/3

=))(1/2)^200>(1/3)^300

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

\(a.\frac{1}{2^{300}}=\frac{1}{\left(2^3\right)^{100}}=\frac{1}{8^{100}}\)

\(\frac{1}{3^{200}}=\frac{1}{\left(3^2\right)^{100}}=\frac{1}{9^{100}}\)

\(\text{Vì }\frac{1}{8}>\frac{1}{9}\Rightarrow\frac{1}{\left(2^3\right)^{100}}>\frac{1}{\left(3^2\right)^{100}}\Rightarrow\frac{1}{2^{300}}>\frac{1}{3^{200}}\)

\(b.\frac{1}{5^{199}}:\text{Giữ nguyên}\)

\(\frac{1}{3^{200}}=\frac{1}{3^{199}\cdot3}\)

\(\frac{1}{5^{199}}< \frac{1}{3^{199}\cdot3}\Rightarrow\frac{1}{5^{199}}< \frac{1}{3^{200}}\)

2 bài dưới bn làm tương tự nhé

Bài làm

Ta có: \(\left(-\frac{1}{4}\right)^2=\left(\frac{1}{4}\right)^2\)

\(\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{4}\right)^2\right]^5=\left(\frac{1}{4}\right)^{10}\)

Mà \(2< 10\)

=> \(\left(\frac{1}{4}\right)^2< \left(\frac{1}{4}\right)^{10}\)

Hay \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)

Vậy \(\left(-\frac{1}{4}\right)^2< \left(\frac{1}{8}\right)^5\)

# Học tốt #

Thank you very much!!!!!!

Ta có

\(A=\frac{\left(1^2-2^2\right)\left(1^2-3^2\right).....\left(1^2-2014^2\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)3\left(-2\right)4.....\left(-2013\right)2015}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left[\left(-1\right)\left(-2\right)...\left(-2013\right)\right]\left(3.4.5...2015\right)}{\left(2.3.4.....2014\right)\left(2.3....2014\right)}\)

\(\Leftrightarrow A=\frac{\left(-1\right)2015}{2014.2}=-\frac{2015}{4028}< -\frac{2014}{4028}=-\frac{1}{2}\)

=> A<-1/2