Ta có: \(8^{34}-8^{33}-8^{32}=8^{32}\left(8^2-8-1\right)\)

\(=8^{32}.55\)

Mà 55 chia hết cho 11 nên \(8^{32}.55⋮11\)

\(\Rightarrow\)đpcm

bạn ơi giải thích

Help me ban eiii

\(60!=1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot59\cdot60=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2\cdot4\cdot6\cdot...\cdot58\cdot60\)

\(=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\cdot1\cdot2\cdot3\cdot...\cdot30=1\cdot3\cdot5\cdot...\cdot57\cdot59\times2^{30}\times30!\)

\(\Rightarrow1\cdot3\cdot5\cdot...\cdot59=\frac{60!}{30!\times2^{30}}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}\)đpcm.

\(\frac{31}{2}\cdot\frac{32}{2}\cdot...\cdot\frac{60}{2}\cdot2\cdot4\cdot...\cdot58\cdot60\)

=31.32.33.34...60.1.2.3.4.5...29.30

=1.2.3.4.5.6.7.8.9.10...60

1.3.5.7...59.2.4.6.8...60

=1.2.3.4.5.6...60

Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}\cdot...\cdot\frac{60}{2}=1\cdot3\cdot5\cdot...\cdot59\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S=\left[\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right]+\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right]+\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right]\)

\(S< \left[\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right]+\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]\)

\(S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}\)

\(S< \frac{37}{60}< \frac{48}{60}=\frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{60}\)\(CMR:S< \frac{4}{5}\)

Số số hạng của S là: (60 - 31 ) + 1 = 30 ( số ), chia thành 6 nhóm, mỗi nhóm 5 số hạng.

Ta có: 

\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}+\frac{1}{31}\)

\(\frac{1}{36}+\frac{1}{37}+\frac{1}{38}+\frac{1}{39}+\frac{1}{40}< \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\)

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+\frac{1}{45}< \frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}+\frac{1}{41}\)

\(\frac{1}{46}+\frac{1}{47}+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}< \frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}+\frac{1}{46}\)

\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+\frac{1}{55}< \frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}+\frac{1}{51}\)

\(\frac{1}{56}+\frac{1}{57}+\frac{1}{58}+\frac{1}{59}+\frac{1}{60}< \frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}+\frac{1}{56}\)

\(=>S=\frac{5}{31}+\frac{5}{36}+\frac{5}{41}+\frac{5}{46}+\frac{5}{51}+\frac{5}{56}\)

\(=>S< 0,78...\)\(=>S< \frac{7}{10}\)( mình ước lượng thôi nha )

Vậy \(S< \frac{4}{5}\)vì \(\frac{4}{5}=\frac{8}{10}< \frac{7}{10}\)

~UMK..., mình ko chắc đúng ko nữa~

a) \(\left(3^{35}+3^{34}-3^{33}\right):3^{32}\)

\(=\frac{3^{35}}{3^{32}}+\frac{3^{34}}{3^{32}}-\frac{3^{33}}{3^{32}}\)

\(=3^3+3^2-3\)

\(=27+9-3\)

\(=33\)

b) \(5^3.37+5^3.64-5^7:5^4\)

\(=5^3.37+5^3.64-5^3\)

\(=5^3\left(37+64-1\right)\)

\(=5^3.100\)

\(=125.100\)

\(=12500\)

\(\left(3^{35}+3^{34}-3^{33}\right)\div3^{32}=3^{33}\left(3^2+3-1\right)\div3^{32}\)

\(=3^{33}.11\div3^{32}=11\left(3^{33-32}\right)=11.3=33\)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

Vậy A > 3/5

Phần 2. 

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50)  < 1/4 ;   (1/51 + 1/52+...+1/59+1/60) < 1/5

Mà S = (1/3 + 1/4 + 1/5) < 4/5 (Vì 1/3 + 1/5 < 3/5 hay 7/12 < 3/5 hay 35/60 < 36/60)

Vậy S <  4/5

Thank You