\(\left(\frac{1}{-2}\right)^{40}=\left(\frac{1}{2}\right)^{40}=\frac{1}{2^{40}}=\frac{1}{\left(2^{10}\right)^4}=\frac{1}{1024^4}<\frac{1}{\left(10^3\right)^4}=\frac{1}{1000^4}=\left(\frac{1}{-10}\right)^{12}\)

Bài 1:

Ta có:

\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)

\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)

Lại có:

\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)

\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)

Bài 2:

Ta có:

\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)

\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)

Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)

\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)

\(\Rightarrow13A>13B\Rightarrow A>B\)

 Theo  thứ tự nhé

a) <

b) <

c) >

làm được bài 1:

TA CÓ: \(\left(\frac{1}{16}\right)^{200}=\left(\frac{1}{16}\right)^{200}\)

            \(\left(\frac{1}{2}\right)^{1000}=\left(\frac{1}{2}\right)^{5.200}=\left(\frac{1^5}{2^5}\right)^{200}=\left(\frac{1}{32}\right)^{200}\)

vì mũ số bằng nhau nên ta so sánh phân số. Vì \(\frac{1}{16}>\frac{1}{32}\)nên \(\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{32}\right)^{200}\)do đó\(\left(\frac{1}{16}\right)^{200}>\left(\frac{1}{2}\right)^{1000}\)

Giúp với :)) !!!!!

Ta có :

\(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(=\frac{12}{4.16}+\frac{20}{16.36}+...+\frac{388}{9216.9604}+\frac{396}{9604.10000}\)

\(=\frac{1}{4}-\frac{1}{16}+\frac{1}{16}-\frac{1}{36}+...+\frac{1}{9604}-\frac{1}{10000}\)

\(=\frac{1}{4}-\frac{1}{10000}< \frac{1}{4}\)

\(\Leftrightarrow B< \frac{1}{4}\)

B=\(\frac{12}{4.16}\)+\(\frac{20}{16.36}\)+...+\(\frac{396}{9604.10000}\)

Ta có:\(\frac{12}{4.16}\)=\(\frac{1}{4}\)-\(\frac{1}{16}\)

         \(\frac{20}{16.36}\)=\(\frac{1}{16}\)-\(\frac{1}{36}\)

            ...

Khi đó:B=\(\frac{1}{4}\)-\(\frac{1}{16}\)+\(\frac{1}{16}\)-\(\frac{1}{36}\)+...+\(\frac{1}{9604}\)-\(\frac{1}{10000}\)=\(\frac{1}{4}\)-\(\frac{1}{10000}\)<\(\frac{1}{4}\)

Vậy: B<\(\frac{1}{4}\)

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