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NN
21 tháng 8 2020
a) \(\frac{3^{17}.81^{11}}{27^{10}.9^{15}}=\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
b) \(\frac{9^2.2^{11}}{16^2.6^3}=\frac{\left(3^2\right)^2.2^{11}}{\left(2^4\right)^2.2^3.3^3}=\frac{3^4.2^{11}}{2^8.2^3.3^3}=\frac{3^4.2^{11}}{2^{11}.3^3}=3\)
c) \(\frac{2^{10}.3^{31}+2^{40}.3^6}{2^{11}.3^{31}+2^{41}.3^6}=\frac{2^{10}.3^{31}+2^{40}.3^6}{2.\left(2^{10}.3^{31}+2^{40}.3^6\right)}=\frac{1}{2}\)
23 tháng 7 2016
\(H=\frac{4}{15}-\frac{23}{28}-\left(-\frac{23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}+\frac{-23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}+\frac{-2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}+\frac{-2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=1+\frac{22}{27}\)
\(H=\frac{27}{27}+\frac{22}{27}=\frac{49}{27}\)
23 tháng 7 2016
\(H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}-\frac{2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}-\frac{2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=\frac{49}{27}\)
27 tháng 7 2018
I don't now
sorry
...................
nha
27 tháng 7 2018
\(A=\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{3^{15}}=9\)
\(B=\frac{45^{10}.5^{10}}{75^{10}}=\frac{5^{10}.3^{20}.5^{10}}{5^{20}.3^{10}}=3^{10}\)
\(C=\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4\right)^5.\left(0,2\right)^5}{0,4^6}=\frac{0,2^5}{0,2^2}=0,2^3\)
\(D=\frac{8^{10}+4^{10}}{8^{11}+4^{11}}=\frac{4^{10}\left(2^{10}+1\right)}{4^{11}\left(2^{11}+1\right)}=\frac{2^{10}+1}{2^{13}+1}\)
XO
3 tháng 8 2020
a) \(\frac{\left(17\frac{2}{9}-15\frac{2}{15}\right):5\frac{2}{9}}{\left(18+3,75\right):0,25}.25\%=\frac{\left(17+\frac{2}{9}-15-\frac{2}{15}\right):\frac{47}{9}}{\left(18+3,75\right).4}.\frac{1}{4}\)
\(=\frac{\left(2+\frac{4}{45}\right).\frac{9}{47}}{\left(18+3,75\right).16}=\frac{\frac{94}{45}.\frac{9}{47}}{288+60}=\frac{\frac{2}{5}}{348}=\frac{2}{5}.\frac{1}{348}=\frac{174}{5}=34,8\)
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{1}{2}\)
3 tháng 8 2020
b) \(\frac{\frac{5}{3}-\frac{5}{7}+\frac{5}{9}}{\frac{10}{3}-\frac{10}{7}+\frac{10}{9}}=\frac{5.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}{10.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{9}\right)}=\frac{5}{10}=\frac{1}{2}\)
NH
7 tháng 7 2019
\(a,\left(x+1\right)^2=81\)
\(\left(x+1\right)^2=9^2\) Hoặc \(\left(x+1\right)^2=\left(-9\right)^2\)
\(\left(x+1\right)=9\) \(x+1=-9\)
\(x=8\) \(x=-10\)
b,\(\left(x+5\right)^{^{ }3}=-64\)
\(\left(x+5\right)^3=\left(-4\right)^3\)
\(x+5=-4\)
=> \(x=-9\)
c,\(\left(2x-3\right)^2=9\)
=>\(\left(2x-3\right)^2=3^2\)Hoặc \(\left(2x-3\right)^2=\left(-3\right)^2\)
\(2x-3=3\) \(2x-3=-3\)
\(2x=6\) \(2x=0\)
=> \(\hept{\begin{cases}x=3\\x=0\end{cases}}\)
d, \(\left(4x+1\right)^3=27\)
\(\left(4x+1\right)^{^{ }3}=3^3\)
\(4x+1=3\)
\(4x=2\)
\(x=\frac{1}{2}\)
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{8^6}{4}=\frac{\left(2^3\right)^6}{2^2}=\frac{2^{18}}{2^2}=2^{16}\)
NH
7 tháng 7 2019
\(D=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{15}+4^{10}}{4^6+4^{11}}=\frac{4^{10}.4^5+4^{10}}{4^6+4^6.4^5}=\frac{4^{10}.\left(4^5+1\right)}{4^6.\left(4^5+1\right)}=\frac{4^{10}}{4^6}=4^4=256\)
phần D trên mk làm sai xin lỗi nha
\(\frac{3^{17}\cdot8^{11}}{27^{10}\cdot9^{15}}=\frac{3^{17}\cdot8^{11}}{3^{30}\cdot9^{15}}=\frac{8^{11}}{3^{13}\cdot9^{15}}=\frac{8^{11}}{3^{43}}\)
sai rồi