310 . 410 - 310 . 49 / 39 . 410
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\(a,4^9:4^4=4^5\\b,17^8:17^5=17^3\\ c,2^{10}:8^2=2^{10}:2^6=2^4\\ d,18^{10}:3^{10}=2^{10}\cdot3^{20}:3^{10}=2^{10}\cdot3^{10}=6^{10}\\ e,27^5:81^3=3^{15}:3^{12}=3^3 \)
\(f,10^6:100=10^6:10^2=10^4\\ g,5^9:25^3=5^9:5^6=5^3\\ h,4^{10}:64^3=4^{10}:4^9=4\\ i,2^{25}:32^4:18^4:9^4=2^{25}:2^{20}:2^4:3^8:3^8=2\)
a)
49:44=45
178:175=173
210:82=210:24=26
1810:310=1510
275:813=315:312=33
b)106:102=104
59:56=53
410:643=4
225:324=225:220=25
184:94=94
\(4^9:4^4\)
=\(4^{9-4}\)
=\(4^5\)
\(17^8:17^5\)
=\(17^{8-5}\)
= \(17^3\)
\(2^{10}:8^2\)
\(=2^{^{ }10}:\left(2^3\right)^2\)
=\(2^{10}:2^6\)
\(=2^4\)
\(18^{10}:3^{10}\)
=\(\left(18:3\right)^{10}\)
=\(6^{10}\)
a) \(4^9:4^4\)
\(=4^{9-4}\)
\(=4^5\)
b) \(17^8:17^5\)
\(=17^{8-5}\)
\(=17^3\)
c) \(2^{10}:8^2\)
\(=2^{10}:\left(2^3\right)^2\)
\(=2^{10-6}\)
\(=2^4\)
d) \(18^{10}:3^{10}\)
\(=\left(18:3\right)^{10}\)
\(=6^{10}\)
e) \(27^5:81^3\)
\(=\left(3^3\right)^5:\left(3^4\right)^3\)
\(=3^{15}:3^{12}\)
\(=3^{15-12}\)
\(=3^3\)
f) \(10^6:100\)
\(=10^6:10^2\)
\(=10^{6-2}\)
\(=10^4\)
g) \(5^9:25^3\)
\(=5^9:\left(5^2\right)^3\)
\(=5^9:5^6\)
\(=5^{9-6}\)
\(=5^3\)
h) \(4^{10}:64^3\)
\(=4^{10}:\left(4^3\right)^3\)
\(=4^{10-9}\)
\(=4\)
i) \(2^{25}:32^4:18^4:9^4\)
\(=\left(2^{25}:2^{20}\right):\left(18^4\cdot9^4\right)\)
\(=2^5:9^8:2^4\)
\(=2:9^8\)
\(=\dfrac{2}{9^8}\)
N=M/300=(72.104)/300=2400(Nu)
Ta có:
\(\left\{{}\begin{matrix}2A+2X=2400\\X-A=380\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=T=410\\G=X=790\end{matrix}\right.\)
=> chọn B
Tử số = 18 x 123 + 9 x 4567 x 2 + 5310 x 6
= 18 x 123 + 9 x 2 x 4567 + 1770 x 3 x 6
= 18 x 123 + 18 x 4567 + 1770 x 18
= 18 x (123 + 4567 + 1770)
= 18 x 6460
Mẫu số = 1 + 4 + 7 + 10 + ... + 49 + 52 + 55 + 58 - 410
Đặt A = 1 + 4 + 7 + 10 + ... + 49 + 52 + 55 + 58
Số số hạng của A là: (58 - 1) : 3 + 1 = 20 (số)
=> A = (1 + 58) x 20 : 2 = 590
=> Mẫu số = 590 - 410 = 180 = 18 x 10
=> tử số/mẫu số = 18 x 6460 / 18 x 10 = 646
\(\frac{3^{10}.4^{10}-3^{10}.4^9}{3^9.4^{10}}\)
\(=\frac{3^{10}.4^9\left(4-1\right)}{3^9.4^{10}}\)
\(=\frac{3.3}{4}\)
\(=\frac{9}{4}\)