a: 27^4=(3^3)^4=3^12<3^18

b: 49^4=(7^2)^4=7^8

c: 9^16=(9^2)^8=81^8>27^8

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

Le Thi My Duyen đừng lih tih, trog câu hỏi tương tự CỐ GẮNG LÊN BẠN NHÉ lm lih tih !!

12¹⁸ = 3¹⁸ . 2³⁶

27⁶ . 16⁹ = 3¹⁸ . 3³⁶

=> 12¹⁸ = 27⁶ . 16⁹ 

so sánh 12^18 và 27^16 *16^9

b: \(2^{27}=8^9\)

\(3^{18}=9^9\)

1. Từ bé đến lớn: \(\frac{3}{10};\frac{8}{15};\frac{9}{6};\frac{13}{5};\frac{16}{3}\)

2. 

a, Ta có: \(\frac{24}{36}=\frac{6}{9}\)  ;  \(\frac{15}{27}=\frac{5}{9}\)

Vì \(\frac{6}{9}>\frac{5}{9}\)nên \(\frac{24}{36}>\frac{15}{27}\)

P/s: Các câu còn lại tương tự :>