Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A= \(\frac{25^3.5^3}{6.5^{10}}\)= \(\frac{\left(5^2\right)^3.5^3}{6.5^{10}}\)= \(\frac{5^6.5^3}{6.5^{10}}\)= \(\frac{5^9}{6.5^{10}}\)= \(\frac{5}{6}\)
B = \(\frac{2^5.6^3}{8^2.9^2}\)= \(\frac{2^5.\left(2.3\right)^3}{\left(2^3\right)^2.\left(3^2\right)^2}\)=\(\frac{2^5.2^3.3^3}{2^6.3^4}\)= \(\frac{2^8.3^3}{2^6.3^4}\)= \(\frac{2^2}{3}\)= \(\frac{4}{3}\)
C = \(\frac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)= \(\frac{5^3.3^3+5.5^2.3^2-5^3}{6^3.3^3+6.6^2.3^2-6^3}\)= \(\frac{5^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)= \(\frac{5^3.\left(1+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)= \(\frac{5^3.9}{6^3.35}\)
=\(\frac{5^3.3^2}{2^3.3^3.7.5}\)
= \(\frac{25}{168}\)
D = \(\frac{\left(7^4-7^3\right)^2}{49^3}\)= \(\frac{[7^3\left(7-1\right)]^2}{\left(7^2\right)^3}\)= \(\frac{7^6.6^2}{7^6}\)= \(36\)
\(a,\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[2^4-4^2\right]\)
\(=\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[16-16\right]\)
\(=\left[2^{17}+16^2\right]\left[9^{15}-3^{15}\right]\cdot0=0\)
\(b,\left[8^{2017}-8^{2015}\right]\cdot\left[8^{2014}\cdot8\right]\)
\(=8^{2015}\left[8^2-1\right]\cdot8^{2015}\)
\(=8^{2015}\cdot63\cdot8^{2015}=8^{4030}\cdot63\)sửa lại câu b , có vấn đề rồi
\(c,\frac{2^8+8^3}{2^5\cdot2^3}=\frac{2^8+\left[2^3\right]^3}{2^5\cdot2^3}=\frac{2^8+2^9}{2^8}=\frac{2^8\left[1+2\right]}{2^8}=3\)
2.a, \(2^6=\left[2^3\right]^2=8^2\)
Mà 8 = 8 nên 82 = 82 hay 26 = 82
b, \(5^3=5\cdot5\cdot5=125\)
\(3^5=3\cdot3\cdot3\cdot3\cdot3=243\)
Mà 125 < 243 nên 53 < 35
c, 26 = [23 ]2 = 82
Mà 8 > 6 nên 82 > 62 hay 26 > 62
d, 7200 = [72 ]100 = 49100
6300 = \(\left[6^3\right]^{100}\)= 216100
Mà 49 < 216 nên 49100 < 216100 hay 7200 < 6300
đăng từng câu nhé bạn
chứ kiểu vậy thì ko có ai giải cho bạn đâu
a) 814=(23)14=23*14=242
1610=(8*2)10=810*210=(23)10*210=230*210=240
Vì 242 > 240 nên 814 > 1610
b) 233=(23)11=811
322=(32)11=911
Vì 811 < 911 nên 233 < 322
a) 8.(-5).(-4).2 = 8.20.2 = 8.40 = 320
b) \(1\frac{3}{7}+\left(-\frac{1}{3}+2\frac{4}{7}\right)\)
\(=1\frac{3}{7}-\frac{1}{3}+2\frac{4}{7}\)
\(=\left(1\frac{3}{7}+2\frac{4}{7}\right)-\frac{1}{3}=\left(1+2\right)+\left(\frac{3}{7}+\frac{4}{7}\right)-\frac{1}{3}=4-\frac{1}{3}=\frac{11}{3}\)
c) \(\frac{8}{5}\cdot\frac{-2}{3}+\frac{-5\cdot5}{3\cdot5}\)
\(=\frac{8}{5}\cdot\frac{-2}{3}+\frac{-25}{15}=\frac{-16}{15}+\frac{-25}{15}=\frac{-41}{15}\)
d) \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{13}{56}\)