\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
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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
\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.2.7^{14}}{10.7^{15}}=\frac{1}{7}\)
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=5:\frac{1}{7}=35\)
Tính :
\(\frac{2.5^{22}-9.2^{21}}{5^{20}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\frac{5^{21}.\left(2.5-9\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)
\(=5:\frac{5.2}{7.10}\)
\(=5:\frac{1}{7}\)
\(=35\)
Ta có :
\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
\(=\dfrac{5.7^6\left(3.7^9-19.7^8\right)}{7^6\left(1+3.7^9\right)}\)
\(=\dfrac{5.7^8\left(3.7-19\right)}{1+3.7^9}\)
\(=\dfrac{5.7^8.2}{1+3.7^9}\)
\(=\dfrac{10.7^8}{1+3.7^8.7}\)
\(=\dfrac{10.7^8}{1+7^8.21}\)