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\(D=\frac{2\cdot8^9\cdot27+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(=\frac{2^{28}\cdot3^3+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\frac{3^3\cdot2^{11}\left(2^{17}+3^6\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\frac{2^{18}+2\cdot3^6}{6^4+3^5}\)
Đúng ko ta.Kết quả hổng đẹp chút nào:(((
\(E=\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\cdot\left(\frac{-5}{12}\right)^3}\)
\(E=\frac{\frac{2^3}{3^3}\cdot\frac{3^2}{4^2}\cdot\left(-1\right)}{\frac{2^2}{5^2}\cdot\frac{\left(-5\right)^3}{12^3}}\)
\(E=\frac{\frac{-2}{3\cdot4}}{\frac{2^2}{5^2}\cdot\frac{-5^3}{2^6\cdot3^3}}=-\frac{\frac{1}{3}}{-\frac{5}{2^4\cdot3^3}}=\frac{2^4\cdot3^2}{5}\)
\(a,=\dfrac{3}{2}-\dfrac{5}{6}:\dfrac{1}{4}+\sqrt{\dfrac{1}{4}-\dfrac{1}{2}}=\dfrac{3}{2}-\dfrac{10}{3}+\sqrt{\dfrac{1}{2}}=-\dfrac{11}{6}+\dfrac{\sqrt{2}}{2}=\dfrac{-33+3\sqrt{2}}{6}\)
\(b,=-\dfrac{4}{3}\cdot\dfrac{9}{2}+\dfrac{13}{12}\cdot\left(-\dfrac{8}{13}\right)=6-\dfrac{2}{3}=\dfrac{16}{3}\\ c,=\dfrac{1}{4}-\left(-\dfrac{1}{6}:4-8\cdot\dfrac{1}{16}\right)=\dfrac{1}{4}-\left(-\dfrac{1}{24}-\dfrac{1}{2}\right)\\ =\dfrac{1}{4}-\dfrac{13}{24}=-\dfrac{7}{24}\\ d,=\dfrac{3^{11}\cdot5^{11}\cdot5^7\cdot3^4}{5^{18}\cdot3^{18}}=\dfrac{1}{3^3}=\dfrac{1}{27}\)
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
b)\(-\frac{7}{3}.\frac{5}{9}+\frac{4}{9}.\left(-\frac{3}{7}\right)+\frac{17}{7}\)
=\(-\frac{35}{27}+\left(-\frac{4}{21}\right)+\frac{17}{7}\)
= \(-\frac{35}{27}+\frac{47}{21}\)
= \(\frac{178}{189}\)
c) \(\frac{117}{13}-\left(\frac{2}{5}+\frac{57}{13}\right)\)
= \(\frac{117}{13}-\frac{311}{65}\)
= \(\frac{274}{65}\)
d) \(\frac{2}{3}-0,25:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{4}:\frac{3}{4}+\frac{5}{8}.4\)
= \(\frac{2}{3}-\frac{1}{3}+\frac{5}{2}\)
= \(\frac{1}{3}+\frac{5}{2}\)
= \(\frac{17}{6}\)
a, \(\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
=\(\dfrac{2\cdot\left(2^3\right)^4\cdot\left(3^3\right)^2+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot\left(3^2\right)^4}\)
=\(\dfrac{2\cdot2^{12}\cdot3^6+2^{11}\cdot3^9}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
=\(\dfrac{2^{11}\cdot3^6\cdot\left(2^2+3^3\right)}{2^{10}\cdot3^7\cdot\left(2^4+5\cdot3\right)}\)
=\(\dfrac{2^{11}\cdot3^6\cdot31}{2^{10}\cdot3^7\cdot31}\)
=\(\dfrac{2}{3}\)
b, \(\dfrac{\dfrac{8}{27}\cdot\dfrac{9}{16}\cdot\left(-1\right)}{\dfrac{4}{25}\cdot\dfrac{-125}{1728}}\)
=\(\dfrac{\dfrac{8\cdot9\cdot\left(-1\right)}{27\cdot16}}{\dfrac{4\cdot\left(-125\right)}{25\cdot1728}}\)
=\(\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}\)
=\(\dfrac{-1}{6}\cdot\dfrac{-432}{5}\)
=\(\dfrac{72}{5}\)
3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50
a; \(\dfrac{5}{6}\) + \(\dfrac{5}{12}\) + \(\dfrac{5}{20}\) + ... + \(\dfrac{5}{132}\)
= 5.(\(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + ..+ \(\dfrac{1}{132}\))
= 5.(\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{11.12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ...+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))
= 5.(\(\dfrac{6}{12}\) - \(\dfrac{1}{12}\))
= 5.\(\dfrac{5}{12}\)
= \(\dfrac{25}{12}\)
\(a,\frac{-2}{3}x=8\)<=> \(x=-12\)
\(b,\frac{1}{4}:x=-3+\frac{3}{4}\)<=>\(\frac{1}{4}:x=\frac{-9}{4}\)<=>\(x=-9\)
\(c,\orbr{\begin{cases}2x-\frac{1}{3}=0\\0.5x+0.25=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{6}\\x=\frac{-3}{4}\end{cases}}\)
a) \(\frac{-2}{3}x+4=12\)
\(\Rightarrow\frac{-2}{3}x=12-4\)
\(\Rightarrow\frac{-2}{3}x=8\)
\(\Rightarrow x=8:\frac{-2}{3}\)
\(\Rightarrow x=-12\)
Vậy x = -12
b) \(\frac{-3}{4}+\frac{1}{4}:x=-3\)
\(\Rightarrow\frac{1}{4}:x=-3-\left(\frac{3}{4}\right)\)
\(\Rightarrow\frac{1}{4}:x=\frac{-9}{4}\)
\(\Rightarrow x=\frac{1}{4}:\frac{-9}{4}\)
\(\Rightarrow x=\frac{-1}{9}\)
Vậy \(x=\frac{-1}{9}\)
c) \(\left(2x-\frac{1}{3}\right)\left(0,5x+0,25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{3}=0\\0,5x+0,25=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{1}{3}\\0,5x=-0,25\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-0,5\end{cases}}\)
Vậy \(x=\frac{1}{6}\)hoặc \(x=-0,5\)
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