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\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)
Câu 1 nhầm đề nha bạn mình sửa:
\(\dfrac{1995.1994-1}{1993.1995+1994}\)
\(=\dfrac{1995.\left(1993+1\right)-1}{1993.1995+1994}\)
\(=\dfrac{1995.1993+1995-1}{1993.1995+1994}\)
\(=\dfrac{1993.1995+1994}{1993.1995+1994}\)
\(=1\)
Câu 2: \(\dfrac{2004.2004+3006}{2005.2005-1003}\)
\(=\dfrac{2004.2004+2004+1002}{\left(2004+1\right).\left(2004+1\right)-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1-1003}\)
\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1002}\)
\(=1\)
Câu 3:\(\dfrac{2010.2011-1}{2009.2011+2010}\)
\(=\dfrac{\left(2009+1\right).2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2011-1}{2009.2011+2010}\)
\(=\dfrac{2009.2011+2010}{2009.2011+2010}\)
= 1
Câu 4:Nhầm để, sửa:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
a) \(\frac{28}{36}=\frac{7}{9}\)
b)\(-\frac{63}{90}=-\frac{7}{10}\)
c)\(\frac{40}{-120}=-\frac{1}{3}\)
Gọi phân số cần tìm là a/b
Ta có
\(\frac{a}{b}=\frac{3}{4}\)
\(\frac{a+60}{b}=\frac{9}{10}\)
\=>\(\frac{a}{b}+\frac{60}{b}=\frac{9}{10}\)
=>\(\frac{3}{4}+\frac{60}{b}=\frac{9}{10}\)
\(\frac{60}{b}=\frac{9}{10}-\frac{3}{4}=\frac{3}{20}=\frac{60}{400}\)
=>b=400 , a=300
a, \(\frac{13.2-13.3}{1-27}\)=\(\frac{13.\left(2-3\right)}{-26}\)=\(\frac{13.\left(-1\right)}{-26}\)=\(\frac{-13}{-26}\)=\(\frac{1}{2}\)
b,\(\frac{15.\left(-3\right)+23.15}{-5+20}\)=\(\frac{15.[\left(-3\right)+23]}{15}\)=\(\frac{15.20}{15}\)=\(\frac{300}{15}\)=20
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