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Bài 2: 

a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)

b: \(5A=5+5^2+...+5^{51}\)

\(\Leftrightarrow4A=5^{51}-1\)

hay \(A=\dfrac{5^{51}-1}{4}\)

Bài 3:

\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)

15 tháng 7 2023

a, ( 3 - 0,6) - ( 7 + 3\(\dfrac{1}{4}\) - \(\dfrac{8}{5}\)) - ( 9 - 2\(\dfrac{1}{4}\))

 = 2,4 - (7 + 3,25 - 1,6) - (9 - 2,25)

= 2,4 - 7 - 3,25 + 1,6 - 9 + 2,25

= (2,4 + 1,6) - (7+ 9) - ( 3,25 - 2,25)

= 4 - 16 - 1

= - 12 - 1

= -13

b, ( - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{0}{8}\)) - (\(\dfrac{5}{6}\) - \(\dfrac{7}{8}\) - 1,4) + ( \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\))

 = - \(\dfrac{5}{8}\) + \(\dfrac{7}{6}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{8}\) + \(\dfrac{7}{5}\) + \(\dfrac{3}{4}\) +  \(\dfrac{5}{3}\) + \(\dfrac{12}{5}\)

= (- \(\dfrac{5}{8}\) + \(\dfrac{7}{8}\)) + (\(\dfrac{7}{6}\) - \(\dfrac{5}{6}\)) + ( \(\dfrac{7}{5}\) + \(\dfrac{12}{5}\)) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)

\(\dfrac{1}{4}\) + \(\dfrac{1}{3}\) + \(\dfrac{19}{5}\) + \(\dfrac{3}{4}\) + \(\dfrac{5}{3}\)

= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + ( \(\dfrac{1}{3}\) + \(\dfrac{5}{3}\)) + \(\dfrac{19}{5}\)

= 1 + 2 + 3,8

= 6,8

12 tháng 8 2018

y mang dấu ( + )

x mang dấu ( - ) 

_Chúc Học Tốt_

4 tháng 8 2015

Bạn tính theo từng phần nha          

23 tháng 9 2020

f) \(\left(1:\frac{1}{7}\right)^2\left[\left(2^2\right)^3:2^5\right]\cdot\frac{1}{49}\)

\(=\left(1\cdot7\right)^2:\left(2^6:2^5\right)\cdot\frac{1}{49}=7^2\cdot\frac{1}{2}\cdot\frac{1}{49}=49\cdot\frac{1}{49}\cdot\frac{1}{2}=\frac{1}{2}\)

g) \(\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{\left(2^2\right)^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot\left(3^2\right)^3+\left(2^3\right)^4\cdot3^5}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^6+3^5\right)}=\frac{2^{12}\left[3^5\left(1-3\right)\right]}{2^{12}\left[3^5\left(3+1\right)\right]}=\frac{2^{12}\cdot3^5\cdot\left(-2\right)}{2^{12}\cdot3^5\cdot4}=\frac{-2}{4}=-\frac{1}{2}\)

23 tháng 9 2020

                                                               Bài giải

\(f,\text{ }\left(1\text{ : }\frac{1}{7}\right)^2\left[\left(2^2\right)^3\text{ : }2^5\right]\cdot\frac{1}{49}\)

\(=7^2\left(2^6\text{ : }2^5\right)\cdot\frac{1}{7^2}\)

\(=2\)

\(g,\text{ }\frac{4^6\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^6}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^5\cdot\left(1-3\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=-\frac{2}{4}=-\frac{1}{2}\)