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\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
\(\left[\left(-20,83\right).0,2+\left(-9,17\right).0,2\right]:\left[2,47.0,5-\left(-3.53\right).0,5\right]\)
\(=\left\{0,2.\left[\left(-20,83\right)+\left(-9,17\right)\right]\right\}:\left[2,47.0,5+3,53.0,5\right]\)
\(=\left[0,2.\left(-30\right)\right]:\left[0,5.\left(2,47+3,53\right)\right]\)
\(=\left(-6\right):\left(0,5.6\right)\)
\(=\left(-6\right):3\)
\(=\left(-2\right)\)
[ (-20,83) . 0,2 + (-9,17) . 0,2 ] : [ 2,47 . 0,5 - (-3.53) . 0,5 ]
= [\(\frac{-2083}{500}\) + \(\frac{-917}{500}\) ] [ \(\frac{247}{200}-\frac{-353}{200}\) ]
= -6 : 3 = -2
\(B=\dfrac{120-\left(-0,5\right)\cdot\left(-40\right)\cdot\left(-5\right)\cdot\left(-0,2\right)\cdot20\cdot0\cdot25}{5+10+15+...+2015}\\ =\dfrac{120-0}{5+10+15+...+2015}\\ =\dfrac{120}{5+10+15+...+2015}\\ ĐặtA=5+10+15+2015\\ A=\left(2015+5\right)\cdot\left[\left(2015-5\right):5+1\right]:2\\ A=407030\\ VậyB=\dfrac{120}{407030}=\dfrac{6}{203515}\)
\(\left[3,5+10.\left(0,4\right)^2\right]:\left[\left(0,5\right)^2-\left(\frac{1}{5}\right)^3+2,758\right]\)
\(=\left(3,5+1,6\right):\left[\frac{1}{4}-\frac{1}{125}+2,758\right]\)
\(=5,1:\left[\frac{125}{500}-\frac{4}{500}+\frac{1379}{500}\right]\)
\(=5,1:3\)
\(=1,7\)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^3} = \frac{{{{\left( { - 2} \right)}^3}}}{{{3^3}}} = \frac{{ - 8}}{{27}};\\{\left( {\frac{{ - 3}}{5}} \right)^2} = \frac{{{{\left( { - 3} \right)}^2}}}{{{5^2}}} = \frac{9}{{25}};\\{\left( { - 0,5} \right)^3} = {\left( {\frac{{ - 1}}{2}} \right)^3} = \frac{{{{\left( { - 1} \right)}^3}}}{{{2^3}}} = \frac{{ - 1}}{8};\\{\left( { - 0,5} \right)^2}=\frac{{{{\left( { - 1} \right)}^2}}}{{{2^2}}} = \frac{{1}}{4};\\\,{\left( {37,57} \right)^0} = 1;\,\\{\left( {3,57} \right)^1} = 3,57.\end{array}\)
Trả lời
120-(-0,5).(-40).(-5).(-0,2).20.0,25/5+10+10+1995
=120-[(-0,5).(0,2)].[(-40).0,25].[20.(-5)]/2020
=120-0,1.(-10).-100/2020
=120-101/2020
=120-101/2020
=19/2020
\(=\frac{-120+\frac{1}{2}.\left(-40\right).\left(-5\right).\frac{-1}{5}.20.\frac{1}{4}}{5+20.1+1995}\)
\(=\frac{-120+1.\left(-1\right).-5.1.5}{5+1995}\)
\(=\frac{120.-1.1.-5.1.5}{2000}\)
\(=\frac{-120.1\left(-5+5\right)}{2000}\)
\(=0\)