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\(\frac{377.733+722}{379.733-744}\)=\(\frac{377+722}{379-744}\)
tích đúng cho mk nhé
gọi biểu thức trên là A.
\(B=\frac{377\cdot733+722}{379\cdot733-744}\)
\(=\frac{377\cdot733+722}{377\cdot733+733\cdot2-744}\)
\(=\frac{377\cdot733+722}{377\cdot733+\left(733\cdot2-744\right)}=\frac{377\cdot733+722}{377\cdot733+722}\)
= 1
\(\Rightarrow A=1+3\cdot\left(-4\right)^3-5\cdot\left(-2\right)^5\)
= 1+ (-192) - (-160)
= -31
\(=\frac{\left(379-2\right).733+722}{379.733-744}=\frac{379.733-1466+722}{379.733-744}=\frac{379.733-744}{379.733-744}=1\)
\(\frac{377.733+722}{379.733-744}\)
<=>\(\frac{277062}{277063}\)
\(\approx0,99\)
\(Q=\dfrac{1}{2011}+\dfrac{2}{2010}+\dfrac{3}{2009}+...+\dfrac{2010}{2}+\dfrac{2011}{1}\)
\(Q=\left(1+\dfrac{2}{2011}\right)\left(1+\dfrac{2}{2010}\right)+\left(1+\dfrac{3}{2009}\right)+...+\left(1+\dfrac{2010}{2}\right)+1\)
\(Q=\dfrac{2012}{2011}+\dfrac{2012}{2010}+\dfrac{2012}{2009}+...+\dfrac{2012}{2}+\dfrac{2012}{2012}\)
\(Q=2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(\Rightarrow\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{2012.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}\right)}=\dfrac{1}{2012}\)
\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)
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:
\(\dfrac{2014.2015-1}{2013.2015+2014}\)
\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)
\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)
\(=1\)
1.
\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}+\frac{1}{2^{100}}\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\left(\frac{1}{2^{100}}+\frac{1}{2^{100}}\right)\)
\(=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{99}}\)
cứ làm như vậy ta được :
\(=1+1=2\)
2. Ta có :
\(\frac{2008+2009}{2009+2010}=\frac{2008}{2009+2010}+\frac{2009}{2009+2010}\)
vì \(\frac{2008}{2009}>\frac{2008}{2009+2010}\); \(\frac{2009}{2010}>\frac{2009}{2009+2010}\)
\(\Rightarrow\frac{2008}{2009}+\frac{2009}{2010}>\frac{2008+2009}{2009+2010}\)
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
\(P=\dfrac{2010\cdot2011-1}{2010^2+2009}+\dfrac{744-379\cdot733}{377\cdot733+722}=\dfrac{2010\cdot2011-2010+2009}{2010^2+2009}+\dfrac{733-379\cdot733+11}{377\cdot733+733-11}=\dfrac{2010\cdot\left(2011-1\right)+2009}{2010^2+2009}+\dfrac{733\cdot\left(1-379\right)+11}{733\cdot\left(377+1\right)-11}=\dfrac{2010^2+2009}{2010^2+2009}+\dfrac{733\cdot\left(-378\right)+11}{733\cdot378-11}=1+\left(-1\right)=0\)