(1/-2)^40>(1/10)12

đáp án là>

Đặt \(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).........\left(\frac{1}{100^2}-1\right)\)

\(\Rightarrow A=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.\frac{1-4^2}{4^2}............\frac{1-100^2}{100}\)

\(\Rightarrow A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}............\frac{-9999}{100^2}\)

\(\Rightarrow A=\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}...............\frac{-99.101}{100^2}\)

\(\Rightarrow A=\frac{-\left(1.2.3.............99\right).\left(3.4.5............101\right)}{\left(2.3.4......100\right).\left(2.3.4.............100\right)}\)

\(\Rightarrow A=\frac{-1.101}{100.2}=\frac{-101}{200}\)

Vậy \(A=\frac{-101}{200}\)

Chúc bn học tốt

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

=>\(\)

=>\(A< -\left(\frac{1.2.3....99}{2.3.4...100}\right)=-\frac{1}{100}\)

Mà \(-\frac{1}{100}>-\frac{1}{2}\)

=>\(A>-\frac{1}{2}\) đúng ko nhỉ

\(\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)

= \(\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}.....\frac{9999}{100.100}\)

= \(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}\)

=\(\frac{1.2.3.4.5.....99}{2.3.4.5.....100}.\frac{3.4.5.6.7.....101}{2.3.4.5.6.....100}\)

=\(\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)

ta có \(\frac{-1}{2}< 0< \frac{101}{200}\Rightarrow\frac{101}{200}>\frac{-1}{2}\)

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

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

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

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

 Theo  thứ tự nhé

a) <

b) <

c) >

Ta có:

\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)..\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{2017^2}-1\right)\)

\(A=\left(-\frac{3}{2^2}\right)\left(\frac{-8}{3^2}\right)\left(\frac{-15}{4^2}\right)...\left(\frac{-\left(1-2017^2\right)}{2017^2}\right)\)
( có 2016 thừa số)

\(A=\frac{3.8.15...\left(1-2017^2\right)}{2^2.3^2.4^2...2017^2}\)

\(A=\frac{\left(1.3\right)\left(2.4\right)...\left(2016.2018\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(2017.2017\right)}\)

\(A=\frac{\left(1.2.3....2016\right)\left(3.4.5....2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)

\(A=\frac{1.2018}{2017.2}\)

\(A=\frac{1009}{2017}\)

Ta có : \(\frac{1009}{2017}>0\) (vì tử và mẫu cùng dấu)

           \(\frac{-1}{2}< 0\) (vì tử và mẫu khác dấu)

Vậy A>B