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B1: để x là số nguyên thì: 5 chia hết cho 2x+1
=> \(2x+1\in U\left(5\right)\)
+> \(2x+1\in\left\{1;-1;5;-5\right\}\)
=> \(x\in\left\{0;-1;2;-3\right\}\)
Ít ước thì là số nguyên tố nhất: 1, chia hết cho chính nó là 1 và 1, cái này tính là 1 ước chăng?
Nhiều ước số nhất : 0, số nào 0 cũng chia hết :>
\(B=\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right)...\left(\frac{1}{100}-1\right)\)
\(B=\frac{-3}{4}.\frac{-8}{9}...\frac{-99}{100}\)
\(B=-\left(\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\right)\)
\(B=-\left(\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{9.11}{10.10}\right)\)
\(B=-\left(\frac{1.2...9}{2.3...10}.\frac{3.4...11}{2.3...10}\right)\)
\(B=-\left(\frac{1}{10}.\frac{11}{2}\right)\)
\(B=\frac{-11}{20}< \frac{-11}{21}\)
Vậy \(B< \frac{-11}{21}\)
\(=\frac{\left(379-2\right).733+722}{379.733-744}=\frac{379.733-1466+722}{379.733-744}=\frac{379.733-744}{379.733-744}=1\)
a) \(\dfrac{-12}{16}=\dfrac{-15}{20}=\dfrac{-18}{24}=\dfrac{-21}{28}=\dfrac{-24}{32}=\dfrac{-27}{36}=\dfrac{-30}{40}=\dfrac{-33}{44}=\dfrac{-36}{48}=\dfrac{-39}{52}\)
\(=\dfrac{-42}{56}=\dfrac{-45}{60}=\dfrac{-48}{64}=\dfrac{-51}{68}=\dfrac{-54}{72}=\dfrac{-57}{76}=\dfrac{-60}{80}=\dfrac{-63}{84}=\dfrac{-66}{88}=\dfrac{-69}{92}\)
\(=\dfrac{-72}{96}=\dfrac{-75}{100}\)
P=\(\dfrac{2010\left(2010+1\right)-1}{2010^2+2009}+\dfrac{744-\left[\left(377+2\right).733\right]}{377.733+722}\)
=\(\dfrac{2010^2+2010-1}{2010^2+2009}+\dfrac{744-\left[377.733+1466\right]}{377.733+722}\)
=\(\dfrac{2010^2+2009}{2010^2+2009}+\dfrac{-722-377.733}{377.733+722}\)
=\(1+\left(-1\right)=0\)
Vậy P=0
\(P=\dfrac{2010\cdot2011-1}{2010^2+2009}+\dfrac{744-379\cdot733}{377\cdot733+722}=\dfrac{2010\cdot2011-2010+2009}{2010^2+2009}+\dfrac{733-379\cdot733+11}{377\cdot733+733-11}=\dfrac{2010\cdot\left(2011-1\right)+2009}{2010^2+2009}+\dfrac{733\cdot\left(1-379\right)+11}{733\cdot\left(377+1\right)-11}=\dfrac{2010^2+2009}{2010^2+2009}+\dfrac{733\cdot\left(-378\right)+11}{733\cdot378-11}=1+\left(-1\right)=0\)
\(\frac{377.733+722}{379.733-744}\)
<=>\(\frac{277062}{277063}\)
\(\approx0,99\)
1. a, \(\frac{6}{7}\)=\(\frac{60}{70}\);\(\frac{11}{10}\)=\(\frac{77}{70}\)
vì \(\frac{60}{70}\)<\(\frac{77}{70}\)nên \(\frac{6}{7}\)<\(\frac{11}{10}\)
b, \(\frac{-5}{17}\)<0<\(\frac{2}{7}\)
c, \(\frac{419}{-723}\)<0<\(\frac{-697}{-313}\)
2.
Ta có :\(\frac{2}{6}\)=\(\frac{20}{60}\);\(\frac{5}{12}\)=\(\frac{25}{60}\);\(\frac{4}{15}\)=\(\frac{16}{60}\);\(\frac{8}{20}\)=\(\frac{24}{60}\);\(\frac{10}{30}\)=\(\frac{20}{60}\)
Vì \(\frac{16}{60}\)<\(\frac{20}{60}\)<\(\frac{24}{60}\)<\(\frac{25}{60}\)nên \(\frac{4}{15}\)<\(\frac{2}{6}\)=\(\frac{10}{30}\)<\(\frac{8}{20}\)<\(\frac{5}{12}\)
Bài 1:
A = \(\dfrac{377.733+722}{397.733-744}\)
A = \(\dfrac{377.733+722}{\left(377+2\right).733-744}\)
A = \(\dfrac{377.733+722}{377.733+2.733-744}\)
A = \(\dfrac{377.733+722}{377.733+1466-744}\)
A = \(\dfrac{377.733+722}{377.733+722}\)
A = 1
Bài 2:
B = \(\dfrac{579.933-944}{577.933+922}\)
B = \(\dfrac{\left(577+2\right).933-944}{577.933+922}\)
B = \(\dfrac{577.933+2.933-944}{577.933+922}\)
B = \(\dfrac{577.933+1866-944}{577.933+922}\)
B = \(\dfrac{577.933+922}{577.933+922}\)
B = 1