43-3 nhân [5 mũ 3 : 5 mũ 2 + 5(-4)-2 mũ 2]
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a) \(\frac{75^3.3^7}{81^4.5^6}=\frac{5^3.3^3.5^3.3^7}{\left(3^4\right)^4.5^6}=\frac{5^6.3^3.3^7}{3^{16}.5^6}=\frac{3^{10}}{3^{16}}=\frac{1}{3^6}=\frac{1}{729}\)
b) \(\frac{6^6.4^2}{3^{12}.2^8}=\frac{2^6.3^6.\left(2^2\right)^2}{3^{12}.2^8}=\frac{2^6.3^6.2^4}{3^{12}.2^8}=\frac{2^{10}.3^6}{3^{12}.2^8}=\frac{2^2.1}{3^6}=\frac{4}{729}\)
c) \(\frac{34^5.2^5}{2^{14}.17^5}=\frac{2^5.17^5.2^5}{2^{14}.17^5}=\frac{2^{10}}{2^{14}}=\frac{1}{2^4}=\frac{1}{16}\)
A=5^10 .9+5^10.7 / 5^9.2^4
A=5^10.(9+7) / 5^9.16
A=5^10.16 /5^9.16
A=5^9.5.16/5^9.16
=>A=5/1
=>A=1
B=2^10.55+2^10. 26 / 2^8 . 27
B=2^10.(55+26) / 2^8 .27
B=2^10.81 / 2^8 .27
B=2^8.2^2.81 / 2^8.27
=>B=2^2.81 / 27
=>B=4.81 / 27
=>B=4.3^3.3 / 3^3
=>B=4.3
=>B=12
Cn phần C thì....mik chưa rõ lém nhưng mak nếu ai cko mik là dzui lém òy......ủng hộ nhoa mina
Do UCLN là \(3^3.5^2\Rightarrow\hept{\begin{cases}a\ge3\\b\ge2\end{cases}}\)
Do BCNN là \(3^4.5^3\Rightarrow\hept{\begin{cases}a=4\\b=3\end{cases}}\) vậy a=4 và b=3
\(a,2^4.38-2^4.37=2^4.\left(38-37\right)=2^4.1=16\\ b,4^2.444446-4^3.111111\\ =4^2.\left(444446-4.111111\right)\\ =4^2.2=16.2=32\\ c,\left(2^9.3+2^9.5\right):2^{12}\\ =2^9.\left(3+5\right):2^{12}=2^9.8:2^{12}=2^9.2^3:2^{12}=2^{9+3-12}=2^0=1\\ d,13^2-\left(5^2.4+2^4.15\right)=13^2-5.4.\left(5+4.3\right)\\ =169-20.17\\ =169-340=-171\)
a) 2^4 * 38 - 2^4 * 37 = 16 * 38 - 16 * 37 = 608 - 592 = 16
b) 4^2 * 444446 - 4^3 * 111111 = 16 * 444446 - 64 * 111111 = 7107136 - 7106944 = 192
c) (2^9 * 3 + 2^9 * 5) / 2^12 = (512 * 3 + 512 * 5) / 4096 = (1536 + 2560) / 4096 = 4096 / 4096 = 1
d) 13^2 - (5^2 * 4 + 2^2 * 15) = 169 - (25 * 4 + 4 * 15) = 169 - (100 + 60) = 169 - 160 = 9
\(4^2\cdot120-4^3\cdot17+4^2\cdot34\)
\(=4^2\left(120-4\cdot17+34\right)\)
\(=16\left(120-68+34\right)\)
\(=16\cdot86=1088\)
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.\left(5^2\right)^4}=7.\frac{5^8}{5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3.3^8.5^2.5^3}{3.5.5^4.3^8}=\frac{5^5}{5^5}=1\)
c) Đề hơi sai roi bạn oi
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2=\frac{1}{100}+\frac{121}{100}=\frac{61}{50}\)
\(=43-3\cdot\left[5-20-4\right]=43-3\cdot\left(-19\right)=100\)