Ta có : 27¹⁵ = (3.9)¹⁵ = 3¹⁵ . 9¹⁵ 

Mà 9¹⁵ > 9¹¹ 

Vậy 9¹¹ < 27¹⁵

27^15 = (3.9)^15 = 3^15 . 9^15

ta thấy 9^15 > 9^11

vậy 9^11 < 27^15

11^2<11^5

4^10>8^6

27^3<9^5

Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

A = \(-1+1+\frac{1}{2}\)

A = \(\frac{1}{2}\)

B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)

B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)

B = \(1+1-\frac{11}{27}\)

B = \(\frac{43}{27}\)

Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)

=> A < B

                       Giải

\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)

\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)

\(\Leftrightarrow A=-1+1+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)

\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)

\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)

\(\Leftrightarrow B=1+\frac{-11}{27}+1\)

\(\Leftrightarrow B=2+\frac{-11}{27}\)

\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)

Từ (1) và (2) suy ra A < B

bằng 28 bn nhé

3³⁹ < 11²¹

3⁹⁹ > 11²¹

3³⁹ < 11²¹

3⁹⁹ > 11²¹

Chúc bạn học tốt!!!

33^9<11^21

33⁹ > 11²¹