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1/ Do A > 1 ; B < 1 nên A > B
2/ Áp dụng a/b > 1 <=> a/b < a+m/b+m ( a,b,m thuộc N*)
Do A > 1 nên A < 20158 + 3 + 1 / 20158 - 2 + 1 = 20158 + 4 / 20158 - 1 = B
=> A < B
d) \(\frac{2}{5}.\frac{-4}{7}+\frac{2}{5}.\frac{-3}{7}+1\frac{2}{5}=\frac{2}{5}.\left(\frac{-4}{7}+\frac{-3}{7}\right)+\frac{7}{5}=\frac{2}{5}.\left(-1\right)+\frac{7}{5}=-\frac{2}{5}+\frac{7}{5}=\frac{5}{5}=1\)
a, \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.9^2.6^{19}-7.2^{29}.27^6}=\frac{5.\left(2^2\right)^{15}.\left(3^2\right)^9-2^2.3^{20}.\left(2^3\right)^9}{5.2^9.\left(2.3\right)^{19}-7.2^{29}.\left(3^3\right)^6}=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)\(=\frac{5.2.2^{29}.3^{18}-2^{29}.3^2.3^{18}}{5.2^{28}.3.3^{18}-7.2.2^{28}.3^{18}}=\frac{\left(5.2-3^2\right).2^{29}.3^{18}}{\left(5.3-7.2\right).2^{28}.3^{18}}=2\)
\(\frac{1}{2^2}>\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
cm tt => đpcm
\(\frac{1}{2^2}<\frac{1}{1.2}=1-\frac{1}{2}\)
cmtt =>...................
ta có A=1/2^2+1/3^2+1/4^2+...+1/9^2
mà 1/2^2>1/2.3=1/2-1/3
1/3^2>1/3.4=1/3-1/4
1/4^2>1/4.5=1/4-1/5
........
1/9^2>1/9.10=1/9-1/10
=> 1/2^2+1/3^2+1/4^2+...+1/9^2>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
=>1/2^2+1/3^2+1/4^2+...+1/9^2>1/2-1/10=2/5
vậy A>2/5 *
ta có 1/2^2<1/1.2=1-1/2
1/3^2<1/2.3=1/2-1/3
1/4^2<1/3.4=1/3-1/4
.......
1/9^2<1/8.9=1/8-1/9
=>1/2^2+1/3^2+1/4^2+...+1/9^2<1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
=>1/2^2+1/3^2+1/4^2+...+1/9^2<1-1/9=8/9
vậy A<8/9 **
từ *,** => 8/9>A>2/5 (đpcm)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(1\right)\)
\(S=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\left(2\right)\)
từ (1) và (2) => đpcm
bạn ơi dòng đầu tiên bạn tách sai rồi theo minh thì không phải thế đâu
Xét \(\frac{1}{n}-\frac{1}{\left(n+1\right)}=\frac{1}{n\left(n+1\right)}<\frac{1}{n^2}=\frac{1}{n.n}<\frac{1}{n.\left(n-1\right)}=\frac{1}{\left(n-1\right)}-\frac{1}{n}\) Vậy ta có:
\(S<\frac{1}{2.1}+\frac{1}{3.2}+...+\frac{1}{9.8}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(I\right)\)
Tương tự ta có: \(S>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(II\right)\)
Từ \(\left(I\right),\left(II\right)\) ta có:
\(\frac{2}{5} (ĐPCM)
T = 150.(1+ 2 + 9 + 8) - (1 + 2 + 9 + 8).50
T = (1 + 2 + 9 + 8).(150 - 50)
T = {(1 + 9) + (2+ 8)].100
T = {10 + 10}.100
T = 20.100
T = 2000
T = 150.(1+ 2 + 9 + 8) - (1 + 2 + 9 + 8).50
T = (1 + 2 + 9 + 8).(150 - 50)
T = [(1 + 9) + (2+ 8)].100
T = (10 + 10).100
T = 20.100
T = 2000
Vậy T = 2000