\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
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24 tháng 1 2022 lúc 15:29
\(P=\dfrac{2.5^{22}-9.5^{20}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
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
Đúng 6
Bình luận (3)
\(P=\dfrac{2.5^{22}-9.5^{20}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\dfrac{5^{20}\left(2.5^2-9\right)}{5^{20}}:\dfrac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)
\(=\left(2.5^2-9\right):\dfrac{5\left(3.7-19\right)}{7.10}\)
\(=\dfrac{7.10\left(2.5^2-9\right)}{5\left(3.7-19\right)}\)
\(=\dfrac{7.2\left(2.5^2-9\right)}{3.7-19}\)
\(=\dfrac{14.41}{21-19}\)
\(=\dfrac{14}{2}\cdot41=7.41=287\)
Đúng 2
Bình luận (0)
Tính : \(\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.2.7^{14}}{10.7^{15}}=\frac{1}{7}\)
Đúng 0
Bình luận (1)
1)Tìm x biết: \(\dfrac{1}{2}-\left(\dfrac{1}{3}-\left(\dfrac{1}{4}-\left(\dfrac{1}{5}-x\right)\right)\right)=1\)
2) Tính B = \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
Đúng 0
Bình luận (0)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}+\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=5:\frac{1}{7}=35\)
Đúng 0
Bình luận (0)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=?\)
Tính :
\(\frac{2.5^{22}-9.2^{21}}{5^{20}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
Thực hiện phép tính: \(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\frac{5^{21}.\left(2.5-9\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)
\(=5:\frac{5.2}{7.10}\)
\(=5:\frac{1}{7}\)
\(=35\)
Thực hiện phép tính: \(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\\ =\frac{5^{21}.\left(10-9\right)}{5^{20}}.\frac{7^{15}.\left(7+3\right)}{5.7^{14}.\left(21-19\right)}\\ =5.\frac{70}{10}\\ =5.7\\ =35\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\frac{5^{20}\left(2.5^2-9.5\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{14}\left(7^2+3.7\right)}\)
\(=\left(50-45\right):\frac{5\left(21-19\right)}{49+21}\)
\(=5.\frac{70}{10}=35\)