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Bạn nên sửa đề thành: \(\frac{54\cdot107-53}{53\cdot107+54}\)
Thì \(=\frac{54\cdot\left(54+53\right)-53}{53\left(53+54\right)+54}=\frac{54^2+53\left(53+1\right)-53}{53^2+54\cdot\left(54-1\right)+54}=\frac{54^2+53^2}{53^2+54^2}=1\)
Chứ nguyên đề thì tính hợp lý sao?
b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)
D = \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)
D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)
Chắc vậy
a) Xét \(\frac{1999.2000-2}{1998.1999+3997}=\frac{1999.\left(1998+2\right)-2}{1998.1999+3997}=\frac{1999.1998+1999.2-2}{1998.1999+3997}=\frac{1999.1998+3996}{1998.1998+3997}\)
=> A < B
A=54.107-53/53.107+54
=(53+1).107-53/53.107+54
=53*107+107-53/53.107+54
=53.107+54/53.107+54
=1
B=135.269-133/134.269+135
=(134+1).269-133/134.269+135
=134.296+296-133/134.269+135
=134.269+136/134.269+135
Vì 134.269+136/134.269+135>1 . Nên B>A
\(A=\frac{54.107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
Vậy A > B
A = \(\dfrac{2008}{2009+2010+2011}+\dfrac{2009}{2009+2010+2011}+\dfrac{2010}{2009+2010+2011}\)
Ta có:
\(\dfrac{2008}{2009}>\dfrac{2008}{2009+2010+2011}\)
\(\dfrac{2009}{2010}>\dfrac{2009}{2009+2010+2011}\)
\(\dfrac{2010}{2011}>\dfrac{2010}{2009+2010+2011}\)
Từ 3 điều trên suy ra : A < B
\(\dfrac{54\times107-53}{53\times107+54}=\dfrac{53\times107+53-53}{53\times107+54}=\dfrac{53\times107}{53\times107+54}< 1\)
\(\dfrac{135\times269-133}{134\times269+135}=\dfrac{134\times269+269-133}{134\times269+135}=\dfrac{134\times269+136}{134\times269+135}>1\)
\(\Rightarrow\dfrac{54\times107-53}{53\times107+54}< \dfrac{135\times269-133}{134\times269+135}\)