BT1: Rút gọn
1) \(\frac{377.733+722}{379.733-744}\)
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\(=\frac{\left(379-2\right).733+722}{379.733-744}=\frac{379.733-1466+722}{379.733-744}=\frac{379.733-744}{379.733-744}=1\)
\(\frac{377.733+722}{379.733-744}\)
<=>\(\frac{277062}{277063}\)
\(\approx0,99\)
\(\frac{377.733+722}{379.733-744}\)=\(\frac{377+722}{379-744}\)
tích đúng cho mk nhé
\(M=\frac{377\cdot733+722}{379\cdot733-744}=\frac{377\cdot733+722}{377\cdot733+2\cdot733-744}=\frac{377\cdot733+722}{377\cdot733+722}=1\)
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
\(P=\dfrac{2010\cdot2011-1}{2010^2+2009}+\dfrac{744-379\cdot733}{377\cdot733+722}=\dfrac{2010\cdot2011-2010+2009}{2010^2+2009}+\dfrac{733-379\cdot733+11}{377\cdot733+733-11}=\dfrac{2010\cdot\left(2011-1\right)+2009}{2010^2+2009}+\dfrac{733\cdot\left(1-379\right)+11}{733\cdot\left(377+1\right)-11}=\dfrac{2010^2+2009}{2010^2+2009}+\dfrac{733\cdot\left(-378\right)+11}{733\cdot378-11}=1+\left(-1\right)=0\)
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