\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=5:\frac{1}{7}=35\)

Bạn vào fx gõ cho rõ ra được không?

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

\(=\frac{5^{21}.\left(2.5-9\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)

\(=5:\frac{5.2}{7.10}\)

\(=5:\frac{1}{7}\)

\(=35\)

gì ghi ra

\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\\ =\frac{5^{21}.\left(10-9\right)}{5^{20}}.\frac{7^{15}.\left(7+3\right)}{5.7^{14}.\left(21-19\right)}\\ =5.\frac{70}{10}\\ =5.7\\ =35\)

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

\(=\frac{5^{20}\left(2.5^2-9.5\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{14}\left(7^2+3.7\right)}\)

\(=\left(50-45\right):\frac{5\left(21-19\right)}{49+21}\)

\(=5.\frac{70}{10}=35\)

ad em giải nhanh nhất 

Bài làm

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

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

\(C=\frac{5^{21}.\left(2-9\right).5}{\left(5.5\right)^{10}}:\frac{5.[7^{15}.\left(3-19\right)].7}{7^{15}.\left(3+1\right).7}\)

\(C=\frac{5^{21}.\left(-7\right).5}{5^{10}.5^{10}}:\frac{5.7^{15}.\left(-16\right).7}{7^{15}.4.7}\)

\(C=\frac{5^{21}.\left(-35\right)}{5^{10}.5^{10}}:\frac{7^{15}.\left(-112\right).5}{7^{15}.28}\)

\(C=5.\left(-35\right):\frac{7^{15}.560}{7^{15}.28}\)

\(C=5.\left(-35\right):\frac{1.560}{1.28}\)

\(C=5.\left(-35\right):20\)

\(C=5.\left(-35\right).\frac{1}{20}\)

\(C=-\frac{175}{20}\)

\(C=-\frac{35}{4}\)

Vậy biểu thức trên \(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)bằng \(C=-\frac{35}{4}\)

# Chúc bạn học tốt #

cô giáo bn chữa bài đó chưa, mình làm đúng hay sai vậy

\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.\left[7^{14}\left(3.7-19\right)\right]}{7^{15}\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{2}{7.2}=\frac{2}{14}=\frac{1}{7}\)

a) \(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{5^{21}.\left(2.5-9\right)}{\left(5^2\right)^{10}}=\frac{5^{21}.1}{5^{20}}=5\)

b) \(\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.7^{14}.\left(3.7-19\right)}{7^{15}.\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{1}{7}\)

a) \(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{2.5^{20}.5^2-9.5^{20}.5}{25^{10}}=\frac{\left(5^2\right)^{10}\left(2.5^2-9.5\right)}{25^{10}}=\frac{25^{10}\left(2.5^2-9.5\right)}{25^{10}}=2.5^2-9.5=5\left(2.5-9\right)=5.\left(10-9\right)=5.1=5\)

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

\(P=\dfrac{2.5^{22}-9.5^{20}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)

\(=\dfrac{5^{20}\left(2.5^2-9\right)}{5^{20}}:\dfrac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)

\(=\left(2.5^2-9\right):\dfrac{5\left(3.7-19\right)}{7.10}\)

\(=\dfrac{7.10\left(2.5^2-9\right)}{5\left(3.7-19\right)}\)

\(=\dfrac{7.2\left(2.5^2-9\right)}{3.7-19}\)

\(=\dfrac{14.41}{21-19}\)

\(=\dfrac{14}{2}\cdot41=7.41=287\)