\(\frac{16}{5}\cdot\frac{15}{64}-\frac65\cdot\frac13\)

\(=\frac{2^4\cdot5\cdot3}{5\cdot2^4\cdot4}-\frac{3\cdot2}{5\cdot3}=\frac34-\frac25\)

Ta có BCNN(4,5)=20

\(\frac34-\frac25=\frac{15}{20}-\frac{8}{20}=\frac{7}{20}\)

x là số nguyên tố hay tự nhiên,nguyên dương hoặc nguyên âm vậy?

a)\(\frac{2003.2004-1}{2003.2004}=\frac{1.\left(2003.2004\right)-1}{2003.2004}\) \(=1-\frac{1}{2003.2004}\)

\(\frac{2004.2005-1}{2004.2005}=\frac{1.\left(2004.2005\right)-1}{2004.2005}=1-\frac{1}{2004.2005}\)

Vì \(1-\frac{1}{2003.2004}<1-\frac{1}{2004.2005}\) nên \(\frac{2003.2004-1}{2003.2004}<\frac{2004.2005-1}{2004.2005}\)

b)\(\) \(\frac{1999.2000}{1999.2000+1}=\frac{1999.2000+1-1}{1999.2000+1}=\frac{1.\left(1999.2000+1\right)-1}{1999.2000+1}\)

\(=1-\frac{1}{1999.2000+1}\)

\(\frac{2000.2001}{2000.2001+1}\) \(=\frac{2000.2001+1-1}{2000.2001+1}=\frac{1.(2000.2001+1)-1}{2000.2001+1}\)

\(=1-\frac{1}{2000.2001+1}\)

Vì \(1-\frac{1}{1999.2000+1}\) \(<1-\frac{1}{2000.2001+1}\) nên \(\frac{1999.2000}{1999.2000+1}<\frac{2000.2001}{2000.2001+1}\)

(có giải thích rồi mong bạn hiểu)

\(\frac{-5}{8}\) \(-\) \(\frac{7}{-12}\)

=\(\frac{-5}{8}-\frac{-7}{12}\) =\(\frac{-30}{48}-\frac{-28}{48}\)

=\(\frac{-2}{48}=\frac{-1}{24}\)

=24-[124+158]+158

=24-124-158+158

=24-124=-100

\(\frac{1}{10}\) \(A\) \(=\) \(\) \(\frac{10^{100}+1}{10^{100}+10}\) \(=\) \(\frac{10^{100}+10-9}{10^{100}+10}=1-\frac{9}{10^{100}+10}\)

\(\frac{1}{10}B\) \(=\) \(\frac{10^{101}+1}{10^{101}+10}=\frac{10^{101}+10-9}{10^{101}+10}=1-\frac{9}{101^{101}+10}\)

Vì \(1-\frac{9}{10^{101}+10}>1-\frac{9}{10^{100}+10}\) nên \(\frac{1}{10}A<\frac{1}{10}B\)

Suy ra \(A

Vậy \(A

=54,47 x 8 + 54,47 x 4 - 54,47 x 2

=54,47 x (8+4-2)

=54,47 x 10 = 544,7

5.(12-x)=30+20=50

12-x=50:5=10

x=12-10=2

Tại sao cả hai bạn có tên giống nhau??????