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A = 1/2! + 2/3! + 3/4! + ... + 2015/2016!
A = 2/2! - 1/2! + 3/3! - 1/3! + 4/4! - 1/4! + ... + 2016/2016! - 1/2016!
A = 1 - 1/2! + 1/2! - 1/3! + 1/3! - 1/4! + ... + 1/2015! - 1/2016!
A = 1 - 1/2016! < 1 (đpcm)
M = 1/52 + 1/62 + 1/72 + ... + 1/1002
M > 1/5.6 + 1/6.7 + 1/7.8 + ... + 1/100.101
M > 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8 + ... + 1/100 - 1/101
M > 1/5 - 1/101 > 1/5 - 1/30 = 1/6 = B
=> M > B (đpcm)
C = 1/20 + 1/21 + 1/22 + ... + 1/200
C > 1/200 + 1/200 + 1/200 + 1/200
(181 phân số 1/200)
C > 1/200 . 181 = 181/200 > 180/200 = 9/10 (đpcm)
\(\frac{3^{11}\div5+3^{11}.3}{3^{10}.2^2}=\frac{3^{11}\cdot\frac{1}{5}+3^{11}\cdot3}{3^{10}.2^2}\)
\(=\frac{3^{11}.\left(\frac{1}{5}+3\right)}{3^{10}.2^2}=\frac{3^{11}.\frac{16}{5}}{3^{10}.2^2}=\frac{\frac{48}{5}}{2^2}=\frac{48}{5}\cdot\frac{1}{4}=\frac{12}{5}\)
\(\frac{1}{3^{10}}\div\frac{1}{9^5}=\frac{1}{3^{10}}\div\frac{1}{3^{10}}=1\)
a)Đặt \(A=8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
\(A< 8^9+8^9+8^9+8^9+8^9+8^9+8^9+8^9\)
\(A< 8\cdot8^9\)
\(A< 8^{10}< 9^{10}\)
\(\Rightarrow9^{10}>8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
a) \(8^9+7^9+6^9+5^9+4^9+3^9+2^9+1^9\)
(8+7+6+5+4+3+2+1)9
369
Vậy369>99
#)Giải :
Câu 2 :
Ta có : \(\frac{1}{9^5}=\frac{1}{\left(3^2\right)^5}=\frac{1}{3^{10}}\)
Vì \(\frac{1}{3^5}=\frac{1}{3^5}\Rightarrow\frac{1}{3^5}=\frac{1}{9^5}\)
311:5+311 .3/310.22(Câu 1)
\(=\frac{8.3^{11}}{3^{10}.4}=6\)
1/310 : 1/95 (Câu 2)
\(=\frac{1}{3^{10}}.3^{10}=1\)
(-25/36)5:5/62 (câu 3)
\(=\frac{-5^5}{6^{10}}.\frac{6^2}{5}=-\frac{5^4}{6^8}\)