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a)\(=\frac{27}{23}+\frac{5}{21}-\frac{4}{23}+\frac{1}{2}\)+\(\frac{16}{21}\)
\(=\left(\frac{27}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)\)+\(\frac{1}{2}\)
\(=1+1+\frac{1}{2}\)
\(=2+\frac{1}{2}\)=\(\frac{5}{2}\)
\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)
\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)
\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)
\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)
\(=3,75.\left(7,2+2,8\right)\)
\(=3,75.10=37,5\)
\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)
\(=\frac{-3}{7}+-\frac{4}{7}=-1\)
\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)
\(=9-\frac{1}{8}.8+0,2\)
\(=9-1+0,2=8+0,2=8,2\)
a) 4x + 1/3 = 3/4
=> 4x = 3/4 - 1/3
=> 4x = 5/12
=> x = 5/12 : 4
=> x = 5/48
b) 1/3 - 2/5 + 3x = 3/4
=> -1/15 + 3x = 3/4
=> 3x = 3/4 + 1/15
=> 3x = 49/60
=> x = 49/ 60 : 3
=> x = 49/180
c) 3(1/2 - x) + 1/3 = 7/6 - x
=> 3/2 - 3x + 1/3 = 7/6 - x
=> 11/6 - 3x = 7/6 - x
=> 11/6 - 7/6 = -x + 3x
=> 2/3 = 2x
=> x = 2/3 : 2
=> x = 1/3
a) \(4x+\frac{1}{3}=\frac{3}{4}\)
\(4x=\frac{3}{4}-\frac{1}{3}\)
\(4x=\frac{9}{12}-\frac{4}{12}\)
\(4x=\frac{5}{12}\)
\(x=\frac{5}{12}:\frac{4}{1}=\frac{5}{12}.\frac{1}{4}\)
\(x=\frac{5}{48}\)
d)
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\frac{6^2}{7^{3^2}}\)
\(=\frac{36}{7^6}\)
\(\left(\frac{7^3\left(7-1\right)}{7^6}\right)^2\)
\(=\left(\frac{6}{7^3}\right)^2\)
\(=\left(\frac{6^2}{7^{3^2}}\right)\)
\(=\frac{36}{7^6}\)
Code : Breacker
\(E=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}\)
\(=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{2.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}\)
\(=\frac{1}{2}.\frac{3}{4}\)
\(=\frac{3}{8}\)