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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
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a) \(\frac{-28}{19}\). \(\frac{\left(-38\right)}{14}\)= 4
b) \(\frac{-21}{16}\). \(\frac{\left(-24\right)}{7}\)= \(\frac{9}{2}\)
c) \(\frac{12}{17}\). \(\frac{\left(-34\right)}{9}\)= \(\frac{-8}{3}\)
d) \(\frac{\left(-15\right)}{4}\): \(\frac{21}{10}\)= \(\frac{-25}{14}\)
e) \(-2\frac{1}{7}\): \(-1\frac{1}{14}\)= 2
nhớ ủng hộ mik nha mn
a.=4
b.=119/24
c.= -8/3
d.= -25/14
Mình chỉ làm đc từng này thôi
\(\frac{45^3.20^4.18^2}{180^5}=\frac{\left(3^2.5\right)^3.\left(2^2x5\right)^4.\left(2x3^2\right)^2}{\left(2^2.3^2.5\right)^5}\)
\(=\frac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}\)\(=\frac{3^{10}.5^7.2^{10}}{2^{10}.3^{10}.5^5}=\frac{5^7}{5^5}=5^2\)
1: \(=\dfrac{2^{20}\cdot3^2+2^{24}}{2^{16}\cdot2^2\cdot5^2}=\dfrac{2^{20}\left(3^2+2^4\right)}{2^{18}\cdot5^2}=4\)
2: \(=\dfrac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
3: \(=\dfrac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{29}\cdot8}{2^2\cdot3^{28}}=2\cdot3=6\)
Bạn vào fx gõ cho rõ ra được không?