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VM
1
HT
13 tháng 10 2019
\(A=\frac{\left(4^2\right)^{17}.64^{36}}{8^{35}.32^{34}}\)
\(A=\frac{\left(2^4\right)^{17}.\left(2^6\right)^{36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=\frac{2^{68}.2^{216}}{2^{105}.2^{170}}=\frac{2^{284}}{2^{275}}=2^9=512\)
22 tháng 3 2017
7)\(\dfrac{-19}{34}\left(\dfrac{17}{19}+\dfrac{49}{18}\right)+\dfrac{49}{18}\left(\dfrac{19}{34}-\dfrac{18}{7}\right)\)
=\(\dfrac{-19}{34}.\dfrac{17}{19}+\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}-\dfrac{18}{7}.\dfrac{49}{18}\)
=\(\dfrac{1}{2}+\left(\dfrac{49}{18}.\dfrac{-19}{34}+\dfrac{49}{18}.\dfrac{19}{34}\right)-7\)
=\(\dfrac{1}{2}+\left[\dfrac{49}{18}\left(\dfrac{-19}{34}+\dfrac{19}{34}\right)\right]-7\)
=\(\dfrac{1}{2}+0-7=\dfrac{-13}{2}\)
8)\(\dfrac{29}{32}\left(\dfrac{41}{36}-\dfrac{32}{58}\right)-\dfrac{41}{36}\left(\dfrac{29}{32}+\dfrac{18}{41}\right)\)
=\(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{29}{32}.\dfrac{32}{58}-\dfrac{41}{36}.\dfrac{29}{32}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(\left(\dfrac{29}{32}.\dfrac{41}{36}-\dfrac{41}{36}\dfrac{29}{32}\right)-\dfrac{29}{32}.\dfrac{32}{58}+\dfrac{18}{41}.\dfrac{41}{36}\)
=\(0-\dfrac{1}{2}+\dfrac{1}{2}=0\)
E2
0
21 tháng 4 2019
a) \(\frac{-1}{2}+\frac{-1}{3}+\frac{-5}{4}\)
\(=\left(\frac{-1}{2}+\frac{-1}{3}\right)+\frac{-5}{4}\)
\(=\frac{-5}{6}+\frac{-5}{4}\)
\(=\frac{-50}{24}=\frac{-25}{12}\)
21 tháng 4 2019
A)\(\frac{-1}{2}+\frac{-1}{3}+\frac{-5}{4}\)
\(=\frac{-6}{12}+\frac{-4}{12}+\frac{-15}{12}\)
\(=\frac{-25}{12}\)
19 tháng 3 2019
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
\(A=\left(4^2\right)^{17}.\frac{64^{36}}{8^{35}.32^{34}}\)
\(A=16^{17}.\frac{64^{2.36}}{\left(2^3\right)^{35}.\left(2^5\right)^{34}}\)
\(A=...6.\frac{...6^{36}}{2^{105}.2^{170}}\)
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