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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
\(=\frac{-120+\frac{1}{2}.\left(-40\right).\left(-5\right).\frac{-1}{5}.20.\frac{1}{4}}{5+20.1+1995}\)
\(=\frac{-120+1.\left(-1\right).-5.1.5}{5+1995}\)
\(=\frac{120.-1.1.-5.1.5}{2000}\)
\(=\frac{-120.1\left(-5+5\right)}{2000}\)
\(=0\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
Bài 1 và Bài 2 dễ, bn có thể tự làm được!
Bài 3:
a) ta có: 1020 = (102)10 = 10010
=> 10010>910
=> 1020>910
b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)
(-3)50 = 350 = (35)10= 24310
=> 12510 < 24310
=> (-5)30 < (-3)50
c) ta có: 648 = (26)8= 248
1612 = ( 24)12 = 248
=> 648 = 1612
d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)
\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)
\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)


\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
Giúp mik với
trước 5h nha
a) \(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+1995}\)
\(=\frac{120-\left[\left(-0,5\right).\left(-0,2\right)\right].\left[\left(-40\right).0,25\right].\left[\left(-5\right).\left(20\right)\right]}{\left(1995+5\right).\left[\left(1995-5\right)\div5+1\right]\div2}\)
\(=\frac{120-0,1.\left(-10\right).\left(-100\right)}{2000.399\div2}\)
\(=\frac{120-100}{1000.399}\)
\(=\frac{1}{19950}\)
b) \(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-2.5.27+3.5.36}{10.2.18-20.2.27+5.2.3.2.36}\)
\(=\frac{5.18-2.5.27+3.5.36}{20.18-20.2.27+20.3.36}\)
\(=\frac{5.\left(18-2.27+3.36\right)}{20.\left(18-2.27+3.36\right)}\)
\(=\frac{1}{4}\)
c) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)
\(=\left(\frac{-1}{2}\right).\left(\frac{-2}{3}\right).\left(\frac{-3}{4}\right)...\left(\frac{-1998}{1999}\right)\)
\(=\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-1998\right)}{2.3.4...1999}\)
\(=\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)
Ta có : 1998 số (-1) mà 1998 là số chẵn
Vậy tích của 1998 số (-1) = 1
\(\Rightarrow\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)
\(=\frac{1}{1999}\)