Tính:
a, 4 . 5^2 - 3 . 2^3
b, 4^8 : 4^5 - ( 1^30 + 17 ) : 3^2
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a) \(\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{7}{15}+\left(\frac{-14}{30}\right)\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{3}{17}-\frac{3}{17}\right)-\left(\frac{4}{3}-\frac{1}{3}\right)-\left(\frac{7}{15}+\frac{14}{30}\right)\)
\(=1+0-1-\left(\frac{7}{15}+\frac{7}{15}\right)\)
= -14/15
b) \(\left(\frac{5}{8}-\frac{4}{13}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left(\frac{-3}{2}\right)+\frac{7}{-15}\)
\(=\frac{5}{8}-\frac{4}{13}+\frac{3}{2}-\frac{5}{8}-\frac{9}{13}+\frac{3}{2}-\frac{7}{15}\)
\(=\left(\frac{5}{8}-\frac{5}{8}\right)-\left(\frac{4}{13}+\frac{9}{13}\right)+\left(\frac{3}{2}+\frac{3}{2}\right)-\frac{7}{15}\)
\(=0-1+3-\frac{7}{15}\)
\(=2-\frac{7}{15}\)
\(=\frac{23}{15}\)
a) 1/5 +3/17-4/3+4/5-3/17+1/3-7/15-14/30
= (1/5+4/5)-4/3+1/3-(7/15+14/30)
=1-1-28/30=-28/30
\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)
1) \(\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\right)\)
\(=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\frac{4}{5}-\frac{3}{17}+\frac{1}{15}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{3}{17}-\frac{3}{17}\right)-\left(\frac{4}{3}-\frac{1}{15}\right)\)
\(=1+0-\frac{19}{15}\)
\(=-\frac{14}{15}\)
A -2/13+-11/26
=-4/26+-11/26
=-15/26
B 3/30+-1/5
=3/30+-6/30
=-3/30=-1/10
C -3/17+(2/3+3/17)
=( -3/17+3/17)+2/3
=2/3
D -5/7+3/4+-1/5+-2/7+1/4
=( -5/7+ -2/7)+(3/4+1/4)+ -1/5
=(-1)+1+ -1/5
=0+-1/5= -1/5
a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)
b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)
c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)
d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
\(a,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\left(-\frac{3}{8}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
= \(\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
= \(\frac{1}{4}+\frac{1}{2}\)
= \(\frac{3}{4}\)
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, \(4.5^2-3.2^3\)
= 100 - 24
= 76
b, \(4^8:4^5-\left(1^{30}+17\right):3^2\)
\(=4^3-\left(1+17\right):3^2\)
\(=64-18:9\)
\(=64-2\)
=62
a) \(4\cdot5^2-3\cdot2^3\)
\(=4\cdot25-3\cdot8\)
\(=100-24\)
\(=76\)
b) \(4^8:4^5-\left(1^{30}+17\right):3^2\)
\(=4^3-\left(1+17\right):9\)
\(=64-18:9\)
\(=64-2\)
\(=62\)