tinh Q=(23/99-25/999-27/9999)(7/2-7/3-7/6)
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M=8/3.2/5.3/8.10.19/92
=(8/3.3/8).(2/5.10).19/92
=1.4.19/2
=4.19/92
=19/23
N=5/7.5/11+5/7.2/11-5/7.14/11
=5/7.(5/11+2/11-14/11)
=5/7. -7/11= -5/11
Q=(1/99+12/999 +123/9999).(1/2-1/3-1/6)
=(1/99+12/999+123/9999).(3/6+ -2/6+ -1/6)
=(1/99+12/999+123/9999). 0
=0
\(\frac{1}{9}\),\(\frac{7}{9}\),\(\frac{5}{90}\),\(\frac{7}{900}\),\(\frac{13}{99}\),\(\frac{21}{99}\),\(\frac{32}{99}\),\(\frac{53}{99}\),\(\frac{12}{990}\),\(\frac{46}{9900}\),\(\frac{123}{999}\),\(\frac{456}{999}\),\(\frac{14234}{9999}\),\(\frac{13}{9999}\),\(\frac{7}{99900}\),\(\frac{230}{99900}\),\(\frac{7}{999}\),\(\frac{33}{9999}\),\(\frac{17}{999000}\),\(\frac{230}{999900}\)
\(A=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{6}{13}\)(( . là dấu nhân nha )
\(A=\frac{5}{9}.\left(\frac{7}{13}+\frac{6}{13}\right)\)
\(A=\frac{5}{9}\)
\(B=\left(\frac{1}{19}+\frac{12}{999}+\frac{123}{9999}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
Mà \(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=0\)
\(\Rightarrow B=0\)
a) 2543 < 2549
7000 > 6999
4271 = 4271
26 513 > 26517
100 000 > 99 999
99 999 > 9999
b) 27 000 < 30 000
8000 > 9000 - 2000
43 000 = 42 000 + 1000
86 005 < 86 050
72100 > 72 099
23 400 = 23 000 + 400
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
\(\dfrac{7}{9}=\dfrac{7}{9}\)
\(\dfrac{77}{99}=\dfrac{77\div11}{99\div11}=\dfrac{7}{9}\)
\(\dfrac{777}{999}=\dfrac{777\div111}{999\div111}=\dfrac{7}{9}\)
\(\dfrac{7777}{9999}=\dfrac{7777\div1111}{9999\div1111}=\dfrac{7}{9}\)
\(\Rightarrow\dfrac{7}{9}=\dfrac{77}{99}=\dfrac{777}{999}=\dfrac{7777}{9999}\)
\(\dfrac{77}{99}\) = \(\dfrac{77:11}{99:11}\) = \(\dfrac{7}{9}\)
\(\dfrac{777}{999}\) = \(\dfrac{777:111}{999:111}\) = \(\dfrac{7}{9}\)
\(\dfrac{7777}{9999}\) = \(\dfrac{7777:1111}{9999:1111}\) = \(\dfrac{7}{9}\)
Từ những lập luận luận trên ta có:
\(\dfrac{7}{9}\) = \(\dfrac{77}{99}\) = \(\dfrac{777}{999}\) = \(\dfrac{7777}{9999}\) (đpcm)
A = \(9999^{999^{99^9}}\)
Vì 999 không chia hết cho 2 nên \(999^{99^9}\) không chia hết cho 2
Vậy \(999^{99^9}\) = 2k + 1
A = 99992k+1
A = (99992)k.9999
A = \(\overline{...1}\)k. 9999
A = \(\overline{..1}\).9999
A = \(\overline{..9}\)
B = vì 8 ⋮ 2 nên \(8^{7^{6^{5^{3^2}}}}\) ⋮ 2
Vậy B = 92k = (92)k = \(\overline{..1}\)k = \(\overline{..1}\)
Ta có (ghi đầu bài vào)
Q=(23/99-25/999-27/9999)*0
Q=0
VẬy.... chú may mắn