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\(E=5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
\(E=5\times4-4\times3+5-0,3\times20\)
\(E=20-12+5-6\)
\(E=6\)
Hok tốt
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
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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=.....
\(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\)
\(=\frac{1\left(2^5+2^6+2^7+2^8\right)}{2^4\left(2^5+2^6+2^7+2^8\right)}\)
\(=\frac{1}{2^4}=\frac{1}{16}\)
Ta có \(\frac{1}{16}< \frac{1}{6}\)
=> \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)
So sánh \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}\) với \(\frac{1}{6}\) ?
Ta có: \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}=\frac{2^5.\left(1+2+2^2+2^3\right)}{2^9.\left(1+2+2^2+2^3\right)}\)
\(=\frac{1}{2^4}=\frac{1}{16}< \frac{1}{6}\)
Vậy \(\frac{2^5+2^6+2^7+2^8}{2^9+2^{10}+2^{11}+2^{12}}< \frac{1}{6}\)
A= \(\frac{\left(5^4-5^3\right)^3}{125^3}=\frac{\left(625-125\right)^3}{1953125}=\frac{125000000}{1953125}=64\)
B=\(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{729}{2916}=\frac{1}{4}\)
lm dùm mình vs
Bài làm
\(5\sqrt{16}-4\sqrt{9}+\sqrt{25}-0,3\sqrt{400}\)
= \(5.4-4.3+5-0,3.20\)
= \(20-12+5-6\)
= \(8+\left(-1\right)\)
= \(7\)
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