b: \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)

\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{11}\cdot3^9}\)

\(=\dfrac{1}{2}\cdot\dfrac{-2}{3}=\dfrac{-1}{3}\)

Sửa đề câu a) phải là (-2020/2021)⁰

Xin lỗi mn

\(a,=\dfrac{3}{4}+\dfrac{15}{6}\cdot\dfrac{1}{5}-\dfrac{1}{36}\cdot9-\dfrac{2020}{2021}=\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{2020}{2021}=1-\dfrac{2020}{2021}=\dfrac{1}{2021}\\ b,=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^9\cdot3^8\left(2+5\right)}=\dfrac{2\left(-2\right)}{3}=-\dfrac{4}{3}\)

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: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)

\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)

\(=\dfrac{2591}{2^6\cdot3^5}\)

help

sao nhieu qua zay

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: \(=2\dfrac{55}{30}=2+\dfrac{11}{6}\)

\(a,\\ =\dfrac{5}{4}-\dfrac{3}{4}=\dfrac{2}{4}=\dfrac{1}{2}\\ b,\\ =\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{10-6}{15}=\dfrac{4}{15}\\ c,\\ =\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{20-6}{24}=\dfrac{14}{24}=\dfrac{7}{12}\\ d,\\ =\dfrac{4}{3}-\dfrac{1}{4}=\dfrac{12-3}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

 cảm ơn chị ạ

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:

 \(\frac{25^{28}+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}=\frac{\frac{25^{32}-1}{25^4-1}}{\frac{25^{32}-1}{25^2-1}}=\frac{25^2-1}{25^4-1}\)

bn làm kiểu g vậy mk ko hiểu

A

C

a -c