\(\left(\frac{1}{243}\right)^9=\left(\frac{1}{3^4}\right)^9=\frac{1}{3^{4.9}}=\frac{1}{3^{36}}\)

\(\left(\frac{1}{83}\right)^{13}\frac{1}{3^{42}}\Rightarrow\left(\frac{1}{81}\right)^{13}

tham khảo nha

http://olm.vn/hoi-dap/question/166511.html

bn nhấn vào câu hỏi tương tự là đc

ta có: \(\left(\frac{1}{243}\right)^9\)>\(\left(\frac{1}{83}\right)^{13}\)

ta co( \(\frac{1}{243}\))⁹=(\(\frac{1}{3}\))⁴⁵⁼(\(\frac{1}{81}\))^11,25<(\(\frac{1}{83}\))¹³

ta co( \(\frac{1}{243}\))⁹=(\(\frac{1}{3}\))⁴⁵⁼(\(\frac{1}{81}\))^11,25<(\(\frac{1}{83}\))¹³

a) \(\left(\frac{1}{243}\right)^9=\left(\frac{1}{3^5}\right)^9=\frac{1}{3^{45}}\)

\(\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{81}\right)^{13}=\left(\frac{1}{3^4}\right)^{13}=\frac{1}{3^{52}}< \frac{1}{3^{45}}=\left(\frac{1}{243}\right)^9\Rightarrow\left(\frac{1}{83}\right)^{13}< \left(\frac{1}{243}\right)^9\)

b) 1990¹⁰ + 1990⁹

= 1990⁹ ( 1990 + 1 )

= 1990⁹ . 1991 < 1991¹⁰ = 1991⁹ . 1991

Vậy 1990¹⁰ + 1990⁹ < 1991¹⁰

Ta thấy:

25^5=(5^2)^5=5^(2*5)=5^10>5^9.

=>1/5^9>1/25^5.

=>-1/5^9<-1/25^5.

=>(-1/5)^9<(-1/25)^5.

Vậy ...

(1/15)^9 < (1/20)^7