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

\(\frac{1}{9}\),\(\frac{7}{9}\),\(\frac{5}{90}\),\(\frac{7}{900}\),\(\frac{13}{99}\),\(\frac{21}{99}\),\(\frac{32}{99}\),\(\frac{53}{99}\),\(\frac{12}{990}\),\(\frac{46}{9900}\),\(\frac{123}{999}\),\(\frac{456}{999}\),\(\frac{14234}{9999}\),\(\frac{13}{9999}\),\(\frac{7}{99900}\),\(\frac{230}{99900}\),\(\frac{7}{999}\),\(\frac{33}{9999}\),\(\frac{17}{999000}\),\(\frac{230}{999900}\)

đi hỏi toán lớp 3 à 

\(A=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{6}{13}\)(( . 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 \(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=0\)

\(\Rightarrow B=0\)

này nói cho mà bít làm gì có cái luận nào k cho học sinh lớp 6 hỏi toán lớp 3 đâu

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)

S=(1+2)+...+2^6(1+2)=3(1+...+2^6)⋮3

           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 = 9999^2k+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 = 9^2k = (9²)^k = \(\overline{..1}\)^k = \(\overline{..1}\)