=3,7*(3,8+8,3-0,4*5) = 3,7*10,1=37,37

= 37.37

a, 2 x 4 x 8 x 0,5 x 0,125 x 0,25 x ( 0,4321 + 0,5679 )

= ( 2 x 0,5 ) x ( 4 x 0,25 ) x ( 0,125 x 8 ) x ( 0,4321 + 0,5679 )

= 1 x 1 x 1 x ( 0,4321 + 0,5679 )

= 1 x ( 0,4321 + 0,5679 ) 

= 1 x 1 = 1

b, 3,7 x 3,8 + 8,3 x 3,7 - 0,4 x 3,7 x 5

= 3,7 x ( 3,8 + 8,3 ) - ( 0,4 x 5 ) x 3,7

= 3,7 x 12,1 - 2 x 3,7

= 3,7 x ( 12,1 - 2 )

= 3,7 x 10,1 = 37,37

2

6

6

2

6

2

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333333333333333333333333333333

p

Ta có :

\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)

\(=\dfrac{5.7^6\left(3.7^9-19.7^8\right)}{7^6\left(1+3.7^9\right)}\)

\(=\dfrac{5.7^8\left(3.7-19\right)}{1+3.7^9}\)

\(=\dfrac{5.7^8.2}{1+3.7^9}\)

\(=\dfrac{10.7^8}{1+3.7^8.7}\)

\(=\dfrac{10.7^8}{1+7^8.21}\)

\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.2.7^{14}}{10.7^{15}}=\frac{1}{7}\)

có thể giải thích rõ hơn được k

Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)

Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)

= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)

Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)

= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)

Từ (1) và (2) suy ra:

A= 5:\(\dfrac{1}{7}\)=5.7=35

Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35

siêng năng:) hs chăm ngoan , hs giỏi 

Viết vậy đúng đó em

A = 5/(3.7) + 5/(7.11) + 5/(11.15) + ... + 5/(2019.2023)

= 5/4 . [4/(3.7) + 4/(7.11) + 4/(11.15) + ... + 4/(2019.2023)]

= 5/4 . (1/3 - 1/7 + 1/7 - 1/11 + 1/11 - 1/15 + ... + 1/2019 - 1/2023)

= 5/4 . (1/3 - 1/2023)

= 5/4 . 2020/6069

= 2525/6069