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a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)
\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)
\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)
b,\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)
\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)
\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)
\(=3^n\cdot9+1-2^n\cdot4+1\)
\(=3^n\cdot10-2^n\cdot5\)
Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)
\(3^n\cdot10⋮10\)
Vậy : ....
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a/ \(\left(\frac{-2}{3}\right)^4:24=\frac{16}{81}:24=\frac{2}{243}\)
b/ \(\left(\frac{3}{4}\right)^3.4^4=\frac{27}{64}.256=108\)
c/ \(\frac{3.0,8^5}{2,4^4}=\frac{3.0,32768}{33,1776}=\frac{0,98304}{33,1776}=\frac{4}{135}\)
d/ \(\frac{3^3-0,9^5}{2,7^4}=\frac{27-0,59049}{53,1441}=\frac{26,40951}{53,1441}=0,4969415231\)
e/\(\left(\frac{-7}{2}\right)^2+\left(\frac{-3}{4}\right)^3.64-\left(\frac{-61}{5}\right)\)
\(=\frac{49}{4}+\frac{-27}{64}.64+\frac{61}{5}\)
\(=12,25-27+12,2\)
\(=-2,55\)
f/ \(\frac{2^4.2^6}{\left(2^5\right)^2}-\frac{2^5.15^3}{6^3.10^2}=\frac{2^{10}}{2^{10}}-\frac{2^5.5^3.3^3}{2^3.3^3.5^2.2^2}\)
\(=1-\frac{2^5.5^3.3^3}{2^5.3^3.5^2}=1-\frac{5}{1}=-4\)
\(\)
chúc bạn học tốt
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
\(\frac{2^5.15^4}{5^4.6^3}=\frac{2^5.\left(3.5\right)^4}{5^4.\left(2.3^3\right)}=\frac{2^5.3^4.5^4}{5^4.2^3.3^3}=\frac{2^5.3^4.5^4}{2^3.3^3.5^4}=2^2.3=4.3=12\)
k cho mik nha
\(\frac{2^5.15^4}{5^4.6^3}\)=\(\frac{2^2.81}{3^3}\)=36