12+12+12+....=42
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B(6) = {0; 6; 12; 18;…}
B(12) = {0; 12; 24;….}
B(42) = {0; 42; 84;…}
BC(6; 12; 42) = {0; 84; 168,…}
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\(12\cdot56+24+42\cdot12\)
\(=12\cdot56+42\cdot12+12\cdot2\)
\(=12\cdot\left(56+42+2\right)\)
\(=12\cdot100\)
\(=1200\)
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g) (15 + 37) + (52 – 37 – 17) = 15 + 37 + 52 – 37 – 17 = 15 + 52 – 17 = 50 h) (38 – 42 +14) - (38 – 42 - 15) = 38 – 42 +14 - 38 + 42 + 15 = 14 + 15 = 29 i) (27 + 65 ) + (346 - 27- 65) = 27 + 65 + 346 - 27- 65 = 346 k) - (27 + 65) - (346 – 27 - 65) = -27 - 65 - 346 + 27 + 65 = -346 l) (42 – 69 + 17) - (42 + 17) = 42 – 69 + 17 - 42 - 17 = – 69 m) 24. (16 - 5) – 16. (24 – 5 ) = 24. 16 - 24 . 5 - 16 . 24 + 16. 5 = - 24 . 5 + 16. 5 = -8 . 5 = -40 n) (-12). 47 – 12 . 52 + (-12) = -12 (47 - 52 + 1) = -12 . (-4) = 48 p) 51 . (69 - 25) – 69. (51 - 25) = 51 . 69 - 51 . 25 – 69. 51 + 69. 25 = - 51 . 25 + 69. 25 = (- 51+ 69).25 = 450 |
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sua de
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
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12+12+12+..=42
=>12.x=42(với xEZ)
=>x=3.5
Gọi ... là x.Ta có :
12 + 12 + 12 + x = 42
=> 12 x 3 + x = 42
=> 42 + x = 42
=> x = 42 - 42
=> x = 0
Vậy x = 0