B = \(\dfrac{120-\left(-0.5\right).\left(-40\right).\left(-5\right).\left(-0.2\right).20.0.25}{5+10+15+...+2015}\)
Những câu hỏi liên quan
(5+10+15+...+1000)\(\left[\dfrac{2}{5}:0.5+2\left(-0.4\right)\right]:\left(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}+...\dfrac{1}{1000}\right)\)
TÍNH NHANH
\(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+1995}\)
Tính nhanh:
\(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+195}\)
Tính
\(A=\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}+1\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}\right)-\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}+1\right)\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
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\(A=\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}+1\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}\right)-\left(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}\right)\left(\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}+1\right)\)
tất nhên là bằng 00000000000000000000000000000000000000
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1. 25dfrac{1}{7}:left(-dfrac{5}{7}right)-15dfrac{1}{7}:left(-dfrac{5}{7}right)+dfrac{4}{5} 3. 2dfrac{2}{3}:left{left[left(3,72-0.02right)dfrac{10}{37}right]:dfrac{5}{6}+2,8right}-dfrac{7}{15}2. left(3+dfrac{4}{5}-dfrac{5}{12}right)left(dfrac{6}{7}-dfrac{3}{5}right)^24.23+3.left(-dfrac{1}{2}right)^2-22.4+left[left(-2right)^2:dfrac{1}{2}right]
Đọc tiếp
1. \(25\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)-15\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)+\dfrac{4}{5}\) 3. \(2\dfrac{2}{3}:\left\{\left[\left(3,72-0.02\right)\dfrac{10}{37}\right]:\dfrac{5}{6}+2,8\right\}-\dfrac{7}{15}\)
2. \(\left(3+\dfrac{4}{5}-\dfrac{5}{12}\right)\left(\dfrac{6}{7}-\dfrac{3}{5}\right)^2\)
4.23+3.\(\left(-\dfrac{1}{2}\right)^2\)-22.4+\(\left[\left(-2\right)^2:\dfrac{1}{2}\right]\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
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tính
a) left[dfrac{0.8divleft(dfrac{4}{5}cdot1025right)}{0.64-1}+dfrac{left(1.08-dfrac{2}{25}right)divdfrac{4}{7}}{left(6dfrac{5}{7}-3dfrac{1}{4}right)cdot2dfrac{2}{17}}+left(1.2cdot0.5right)divdfrac{4}{5}right]
b) left(0.2right)^{-3}left[left(-dfrac{1}{5}right)^{-2}right]^{-1}+left[left(dfrac{1}{2}right)^{-3}right]^{-2}divleft(2^{-3}right)^{-1}-left(0.175right)^{-2}
c) 2+dfrac{1}{1+dfrac{1}{2+dfrac{1}{1+dfrac{1}{2}}}}
d) dfrac{1}{90}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-dfrac{1}{3}
e) l...
Đọc tiếp
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
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d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
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f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
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tính nhanh
\(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+10+1995}\)
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
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\(=\frac{-120+\frac{1}{2}.\left(-40\right).\left(-5\right).\frac{-1}{5}.20.\frac{1}{4}}{5+20.1+1995}\)
\(=\frac{-120+1.\left(-1\right).-5.1.5}{5+1995}\)
\(=\frac{120.-1.1.-5.1.5}{2000}\)
\(=\frac{-120.1\left(-5+5\right)}{2000}\)
\(=0\)
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Tính nhanh
a)\(\frac{120-\left(-0,5\right).\left(-40\right).\left(-5\right).\left(-0,2\right).20.0,25}{5+10+15+...+1995}\)
b)\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
c)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{1999}-1\right)\)
\(\frac{5.18-10.27+15.36}{10.36-20.54+30.72}\)
\(=\frac{5.18-10.27+15.36}{5.2.18.2-10.2.27.2+15.2.36.2}\)
\(=\frac{5.18-10.27+15.36}{5.8.2.2-10.27.2.2+15.36.2.2}\)
\(=\frac{1}{2.2-2.2+2.2}\)
\(=\frac{1}{2.2}=\frac{1}{4}\)
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
\(\Rightarrow\frac{\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)}{1.1.1...1999}\)
\(=\frac{1}{1999}\)