a) \(\dfrac{22}{55}=\dfrac{2}{5}\)

b) \(\dfrac{-63}{81}=\dfrac{-7}{9}\)

c) \(\dfrac{2.14}{7.8}=\dfrac{2.7.2}{7.2.2.2}=\dfrac{1}{2}\)

d) \(\dfrac{49+7.49}{49}=\dfrac{49.\left(7+1\right)}{49}=\dfrac{49.8}{49}=8\)

\(\dfrac{22}{55}=\dfrac{2}{5}\)                                          \(\dfrac{-63}{81}=\dfrac{-7}{9}\)

\(\dfrac{2.14}{7.8}=\dfrac{28}{56}=\dfrac{1}{2}\)                             \(\dfrac{49+7.49}{49}=8\)

Giải:

a) \(\dfrac{5^3.90.4^3}{25^2.3^2.2^{13}}\) 

\(=\dfrac{5^3.5.3^2.2.\left(2^2\right)^3}{\left(5^2\right)^2.3^2.2^{13}}\) 

\(=\dfrac{5^4.3^2.2.2^6}{5^4.3^2.2^{13}}\) 

\(=\dfrac{5^4.3^2.2^7}{5^4.3^2.2^{13}}\) 

\(=\dfrac{1}{2^6}=\dfrac{1}{64}\) 

b) \(\dfrac{15^2.16^4-15^3.16^3}{12^2.20^3-20^2.12^3}\) 

\(=\dfrac{15^2.16^3.16-15^2.16^3.15}{12^2.20^2.20-20^2.12^2.12}\) 

\(=\dfrac{15^2.16^3.\left(16-15\right)}{12^2.20^2.\left(20-12\right)}\) 

\(=\dfrac{\left(3.5\right)^2.\left(2^4\right)^3.1}{\left(3.2^2\right)^2.\left(2^2.5\right)^2.8}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.\left(2^2\right)^2.\left(2^2\right)^2.5^2.2^3}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^4.2^4.2^3}\) 

\(=\dfrac{3^2.5^2.2^{12}}{3^2.5^2.2^{11}}\) 

\(=2\) 

c) \(\dfrac{2.3+4.6+14.21}{3.5+6.10+21.35}\) 

\(=\dfrac{2.3+2.3.2.2+2.3.7.7}{3.5+3.5.2.2+3.5.7.7}\) 

\(=\dfrac{2.3+2.3.4+2.3.49}{3.5+3.5.4+3.5.49}\) 

\(=\dfrac{2.3.\left(1+4+49\right)}{3.5.\left(1+4+49\right)}\) 

\(=\dfrac{2.3}{3.5}\) 

\(=\dfrac{2}{5}\) 

Chúc bạn học tốt!

a) `(-63)/72 = (-63:9)/(72:9)=(-7)/8`

b) `20/(-140) = (20:20)/(-140:20) = 1/(-7)=(-1)/7`

c) `(3.10)/(5.24) =(3.2.5)/(5.2 .3.4)=1/4`

d) `(6.5-6.2)/(6+3)=18/9=2`

cảm ơn bạn rất nhiều !!!!

a. \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{3}{160}\)

b. \(\dfrac{2\cdot14}{7\cdot8}=\dfrac{1\cdot2}{1\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)

c. \(\dfrac{11\cdot4-11}{2-13}=\dfrac{11\left(4-1\right)}{-11}=\dfrac{1\cdot3}{-1}=-3\)

d. \(\dfrac{49+7\cdot49}{49}=\dfrac{49\cdot\left(1+7\right)}{49}=\dfrac{8}{1}=8\)

Bài 1:

a) \(a=2\cdot3\cdot5\cdot43\)

\(b=7200=2^5\cdot3^2\cdot5^2\)

\(c-4680=2^3\cdot3^2\cdot5\cdot13\)

b) \(\dfrac{8440}{5910}=\dfrac{8440:10}{5910:10}=\dfrac{844}{591}\)

\(\dfrac{1245}{3450}=\dfrac{1245:15}{3450:15}=\dfrac{83}{230}\)

Bài 2:

a) Ước nguyên tố của 140 là:

\(ƯNT\left(140\right)=\left\{2;5;7\right\}\)

Ước nguyên tố của 138 là:
\(ƯNT\left(138\right)=\left\{3;23;2\right\}\)

b) \(A=\dfrac{2^{10}+4^6}{8^4}\)

\(A=\dfrac{2^{10}+2^{12}}{2^{12}}\)

\(A=\dfrac{2^{10}\cdot\left(1+2^2\right)}{2^{12}}\)

\(A=\dfrac{1+4}{2^2}\)

\(A=\dfrac{5}{4}\)

\(B=\dfrac{6^{10}+15\cdot2^{10}\cdot3^9}{12\cdot8^3\cdot27^3}\)

\(B=\dfrac{2^{10}\cdot3^{10}+5\cdot2^{10}\cdot3^{10}}{2^{11}\cdot3^{10}}\)

\(B=\dfrac{2^{10}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{10}}\)

\(B=\dfrac{1+5}{2}\)

\(B=3\)

a: \(=\dfrac{13\cdot\left(9-2\right)}{13}=7\)

b: \(=\dfrac{14\left(3-8\right)}{7\left(1+3\cdot3\right)}=2\cdot\dfrac{-5}{10}=-1\)

c: \(=\dfrac{54-72}{36}=\dfrac{-18}{36}=-\dfrac{1}{2}\)

d: \(=\dfrac{5^3}{10^2\cdot5}=\dfrac{5^2}{100}=\dfrac{1}{4}\)

a) \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{0,3\cdot10}{16\cdot10}=\dfrac{3}{160}\)

b) \(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\cdot\left(5-1\right)}{-17}=\dfrac{1\cdot4}{-1}=-4\)

a)

\(\dfrac{8\cdot5-8\cdot2}{16}=\dfrac{8\left(5-2\right)}{16}=\dfrac{3}{2}\)

b)

\(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=\dfrac{4}{-1}=-4\)

a) \(\dfrac{{50}}{{85}}\) 

Ta có: \(50 =2.5^2; 85= 5.17\)

Thừa số nguyên tố chung là 5 với số mũ nhỏ nhất là 1 nên ƯCLN(50, 85) = 5. Do đó, \(\dfrac{{50}}{{85}}\) chưa là phân số tối giản

Ta có: \(\dfrac{{50}}{{85}} = \dfrac{{50:5}}{{85:5}} = \dfrac{{10}}{{17}}\)

b)\(\dfrac{{23}}{{81}}\)

Ta có: \(23 = 23; 81 = 3^4\)

Chúng không có thừa số nguyên tố chung nên ƯCLN(23, 81) = 1. Do đó, \(\dfrac{{23}}{{81}}\) là phân số tối giản.

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