2.30)

a) \(30=2.3.5\)

\(45=3^2.5\)

\(ƯCLN\left(30;45\right)=3.5=15\)

\(\RightarrowƯC\left(30;45\right)=Ư\left(15\right)=\left\{1;3;5;15\right\}\)

b) \(42=2.3.7\)

\(70=2.5.7\)

\(ƯCLN\left(42;70\right)=2.7=14\)

\(\RightarrowƯC\left(42;70\right)=Ư\left(14\right)=\left\{1;2;7;14\right\}\)

2.31)

a) \(40=2^3.5\)

\(70=2.5.7\)

\(ƯCLN\left(40;70\right)=2.5=10\)

b) \(55=5.11\)

\(77=7.11\)

\(ƯCLN\left(55;77\right)=11\)

2.32)

a) \(2^2.5\)

\(2.3.5\)

\(ƯCLN\left(2^2.5;2.3.5\right)=2.5=10\)

b) \(ƯCLN\left(2^4.3;2^2.3^2.5;2^4.11\right)=2^2=4\)

2.33)

a) \(72=2^3.3^2\)

\(96=2^5.3\)

b) \(ƯCLN\left(a;b\right)=2^3.3=24\)

\(ƯC\left(a;b\right)=Ư\left(24\right)=\left\{1;2;3;4;6;8;12;24\right\}\)

Bài 3

a) \(\dfrac{31}{23}-\left(\dfrac{7}{32}+\dfrac{8}{23}\right)\)

\(=\dfrac{31}{23}-\dfrac{7}{32}-\dfrac{8}{23}\)

\(=\left(\dfrac{31}{23}-\dfrac{8}{23}\right)-\dfrac{7}{32}\)

\(=1-\dfrac{7}{32}\)

\(=\dfrac{25}{32}\)

b) \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)

\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{11}\)

\(=\left(\dfrac{12}{67}-\dfrac{79}{67}\right)+\left(\dfrac{13}{41}+\dfrac{28}{41}\right)+\dfrac{1}{3}\)

\(=-1+1+\dfrac{1}{3}\)

\(=0+\dfrac{1}{3}\)

\(=\dfrac{1}{3}\)

c) \(\left(-30,75\right)+\left(\dfrac{31}{10}-69,25\right)-\left(-6,9\right)\)

\(=\left(-30,75\right)+\dfrac{31}{10}-69,25+6,9\)

\(=\left[\left(-30,75\right)-69,25\right]+\left(3,1+6,9\right)\)

\(=\left(-100\right)+10\)

\(=-90\)

d) \(\left(-34,5\right).\dfrac{11}{25}-65,5\dfrac{11}{25}\)

\(=\dfrac{11}{25}.\left(-34,5-65,5\right)\)

\(=\dfrac{11}{25}.\left(-100\right)\)

\(=-44\)

e) \(\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{5}{4}+\left(-\dfrac{4}{5}+\dfrac{4}{7}\right):\dfrac{5}{4}\)

\(=\left(-\dfrac{1}{5}+\dfrac{3}{7}-\dfrac{4}{5}+\dfrac{4}{7}\right):\dfrac{5}{4}\)

\(=\left[\left(-\dfrac{1}{5}-\dfrac{4}{5}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{5}{4}\)

\(=\left(-1+1\right):\dfrac{5}{4}\)

\(=0:\dfrac{5}{4}\)

\(=0\)

f) \(\left(-\dfrac{2}{3}+\dfrac{3}{13}\right):\dfrac{7}{8}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right):\dfrac{7}{8}\)

\(=\left(-\dfrac{2}{3}+\dfrac{3}{13}-\dfrac{1}{3}+\dfrac{10}{13}\right):\dfrac{7}{8}\)

\(=\left[\left(-\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{3}{13}+\dfrac{10}{13}\right)\right]:\dfrac{7}{8}\)

\(=\left(-1+1\right):\dfrac{7}{8}\)

\(=0:\dfrac{7}{8}\)

\(=0\)

a) 9/25 = 36/100 = 0,36

b) 7/4 = 175/100 = 1,75

c) 56/200 = 28/100 = 0,28

d) 486/300 = 162/100 = 1,62

(x³ - 8y³) : (x - 2y)

= [(x - 2y)(x² + 2xy + 4)] : (x - 2y)

= x² + 2xy + 4

2 3/5 + 1 2/5 × 3 1/2

= 13/5 + 7/5 × 7/2

= 13/5 + 49/10

= 26/10 + 49/10

= 75/10

= 15/2

Bài 3

\(0,15=\dfrac{15}{100}\)

\(4,05=\dfrac{405}{100}\)

\(47,07=\dfrac{4707}{100}\)

\(19,2=\dfrac{192}{100}\)

\(58,148=\dfrac{58148}{1000}\)

a + b = 80

a - b = 64

a \(\times\) b = 576

a : b = 9

Số tiền chưa khuyến mãi:

\(35000.8+20000.6=400000\) (đồng)

Số tiền thực trả:

\(400000-10000=390000\) (đồng)

\(4,9=\dfrac{49}{10}\)

\(1,35=\dfrac{135}{100}\)

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