1. Tính:
a, (1/2)15.(1/4)20 b, (1/9)25 : (1/3)30
2. Rút gọn: A = ( 45. 94 - 2. 69 ) / ( 210. 38 + 68. 20 )
3. Biểu diễn số thập phân dưới dạng phân số và ngược lại:
a, 7/33 b, 7/22 c, 0,(21) d, 0,5(16)
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b: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{11}\cdot3^9}\)
\(=\dfrac{1}{2}\cdot\dfrac{-2}{3}=\dfrac{-1}{3}\)
a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
Bài 1:
a: 6/9=2/3
6/24=1/4
48/96=1/2
42/98=3/7
b: 24/36=2/3
18/30=3/5
15/120=1/8
80/240=1/3
c: 5/25=1/5
75/100=3/4
64/720=4/45
16/1000=2/125
Bài 2:
Các phân số bằng 2/3 là 34/51; 20/30; 84/126
Bài 1 :
a) 6/9 = 2/3 ; 6/24 = 1/4 ; 48/96 = 1/2 ; 42/98 = 3/7
b) 24/36 = 2/3 ; 18/30 = 3/5 ; 15/120 = 1/8 ; 80 / 240 = 1/3
c) 5 / 25 = 1/5 ; 75/100=3/4 ; 64/720=4/45 ; 16/1000=2/125
Bài 2 :
Các phân số bằng 2/3 là :
\(\text{34/51; 20/30; 84/126}\)
a: \(1\dfrac{3}{4}=1+\dfrac{3}{4}\)
b: \(3\dfrac{5}{9}=3+\dfrac{5}{9}\)
c: \(1\dfrac{9}{10}=1+\dfrac{9}{10}\)
d: \(3\dfrac{25}{20}=3+\dfrac{5}{4}\)
e: \(=2\dfrac{55}{30}=2+\dfrac{11}{6}\)
\(a,\\ =\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\ b,\\ =\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\\ c,\\ =\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{20-6}{24}=\dfrac{14}{24}=\dfrac{7}{12}\\ d,\\ =\dfrac{4}{3}-\dfrac{1}{4}=\dfrac{12-3}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)
\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)
\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)
\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)
\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)
\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)
Bài 1:
a) \(\frac{5^2.6^{11}.16^2+6^2.11^6.15^2}{2.6^{12}.10^{14}-81^2.960^3}=\frac{5^2.2^{19}.3^{11}+2^2.3^4.11^6.5^2}{2^{27}.3^{12}.5^{14}-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.2^2.3^4.\left(2^{17}.3^7+11^6\right)}{2^{18}.3^{11}.5^3.\left(2^9.3.5^{11}-1\right)}=\frac{2^{17}.3^7+11^6}{2^{16}.3^7.5.\left(2^9.3.5^{11}-1\right)}\)
b) Đặt A = 2528 + 2524 +....+ 254 +1
=> 254.A = 2532 + 2528 +...+ 258 + 254
=> 254.A - A = 2532 -1
\(A=\frac{25^{32}-1}{25^4-1}\)
tương tự....
Thay vào:
\(\frac{25^{28}+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}=\frac{\frac{25^{32}-1}{25^4-1}}{\frac{25^{32}-1}{25^2-1}}=\frac{25^2-1}{25^4-1}\)