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a)
57 . 257 và 6255
Ta có:
57 . 257 = 1257 =( 53 )7 = 53.7 = 521
6255 = ( 54 )5 =54.5 = 520
Vì 521 > 520 => 57 . 257 > 6255
Vậy 57 . 257 > 6255
b) ( 34 )4 và ( 32 )6
Ta có:
( 34 )4 = 34.4 = 316
( 32 )6 = 32.6 = 312
Vì 316 > 312 => ( 34 )4 > ( 32 )6
Vậy ( 34 )4 > ( 32 )6
c) 8 . 24 và 29
Ta có:
8 . 24 = 23 . 24 = 27
Vì 27 < 29 => 8 . 24 < 29
Vậy 8 . 24 < 29
Học tốt!!!
\(\frac{3^6.\left(3^2.5\right)^4-\left(3.5\right)^3:5^9}{\left(3^3\right)^4.\left(5^2\right)^3+\left(5.3^2\right)^6}\) = \(\frac{3^{14}.5^4-3^3:5^6}{3^{12}.5^6+5^6.3^{12}}\) = \(\frac{3^{14}.5^{10}-3^3}{3^{12}.5^{12}+5^{12}.3^{ }^{12}^{ }}=\frac{3^{11}.5^{10}-1}{2.3^9.5^{12}}\)
\(E=\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)
\(E=\frac{3^6.3^8.5^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)
\(E=\frac{3^{14}.5^4-3^{13}.5^4}{3^{12}.5^6\left(1+1\right)}\)
\(E=\frac{3^{13}.5^4\left(3-1\right)}{3^{12}.5^6.2}\)
\(E=\frac{3}{25}\)
Tính:
\(\dfrac{3^6.45^4-15^{13}.\dfrac{1}{5^9}}{27^4.25^3+45^6}\)
\(=\dfrac{3^6.\left(3^2.5\right)^4-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}\)
\(=\dfrac{3^6.3^8.5^4-3^{13}.5^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)
\(=\dfrac{3^{14}.5^4-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}\)
\(=\dfrac{3^{13}.5^4.\left(3-1\right)}{2.3^{12}.5^6}\)
\(=\dfrac{3.2}{2.5^2}\)
\(=\dfrac{3}{25}.\)
Đoàn Đức Hiếu, Nguyễn Thanh Hằng, Nguyễn Huy Tú, Tuấn Anh Phan Nguyễn, Hoàng Ngọc Anh, ...
Gọi kết quả của phép tính là A
=> \(\dfrac{4}{7.25}\) + \(\dfrac{5}{25.10}\) + \(\dfrac{6}{10.36}\) + \(\dfrac{9}{36.15}\) = A
<=> \(\dfrac{4}{175}\) + \(\dfrac{5}{250}\) + \(\dfrac{6}{360}\) + \(\dfrac{9}{540}\) = A
<=> (\(\dfrac{4}{175}\) + \(\dfrac{1}{50}\)) + (\(\dfrac{1}{60}\) + \(\dfrac{1}{60}\)) = A
<=> (\(\dfrac{200}{8750}\) + \(\dfrac{175}{8750}\)) + \(\dfrac{2}{60}\) = A
<=> \(\dfrac{375}{8750}\) + \(\dfrac{1}{30}\) = A
<=> \(\dfrac{3}{70}\) + \(\dfrac{1}{30}\) = A
<=> \(\dfrac{90}{2100}\) + \(\dfrac{70}{2100}\) = A
<=> \(\dfrac{160}{2100}\)
\(\frac{1}{4}C=\frac{9}{16\cdot25}+\frac{3}{25\cdot28}+\frac{6}{28\cdot34}+\frac{6}{34\cdot40}\)
\(\frac{1}{4}C=\frac{1}{16}-\frac{1}{25}+\frac{1}{25}-\frac{1}{28}+\frac{1}{28}-\frac{1}{34}+\frac{1}{34}-\frac{1}{40}\)
\(\frac{1}{4}C=\frac{1}{16}-\frac{1}{40}\)
\(C=\frac{3}{20}\)