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ta có:\(\frac{1}{2}=\frac{64}{128}>\frac{63}{128}\)
\(\frac{1}{2}=\frac{2008}{4016}< \frac{2008}{2009}\)
=> \(\frac{63}{128}< \frac{1}{2}< \frac{2008}{2009}\)
Vậy \(\frac{63}{128}< \frac{2008}{2009}\)
\(\frac{2009.2009+2008}{2009.2009+2009}=\frac{2009.2009+2009}{2009.2009+2009}-\frac{1}{2009.2009+2009}=1-\frac{1}{2009.2009+2009}\)
\(\frac{2009.2009+2009}{2009.2009+2010}=\frac{2009.2009+2010}{2009.2009+2010}-\frac{1}{2009.2009+2010}=1-\frac{1}{2009.2009+2010}\)
\(\text{Vì }2009.2009+2009<2009.2009+2010\text{ nên: }\frac{1}{2009.2009+2009}>\frac{1}{2009.2009+2010}\)
\(\text{Hay }1-\frac{1}{2009.2009+2009}<\frac{1}{2009.2009+2010}\)
\(\text{Vậy }\frac{2009.2009+2008}{2009.2009+2009}<\frac{2009.2009+2009}{2009.2009+2010}\)
\(\frac{2009.2009+2009}{2009.2009+2010}=\frac{2009.2009+2008+1}{2009.2009+2009+1}\)
Đặt 2009.2009+2008 là a; 2009.2009+2009 là b. Ta so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+1}\)
Qui đồng mẫu số 2 phân số trên
\(\frac{a}{b}=\frac{a\left(b+1\right)}{b\left(b+1\right)}=\frac{a.b+a}{b.\left(b+1\right)}\)
\(\frac{a+1}{b+1}=\frac{\left(a+1\right).b}{b\left(b+1\right)}=\frac{a.b+b}{b\left(b+1\right)}\)
Vì 2008 < 2009
=> 2009.2009+2008 < 2009.2009+2009
=> a < b
=> a.b+a < a.b+b
=> \(\frac{a.b+a}{b.\left(b+1\right)}<\frac{a.b+b}{b.\left(b+1\right)}\)
=> \(\frac{a}{b}<\frac{a+1}{b+1}\)
=> \(\frac{2009.2009+2008}{2009.2009+2009}<\frac{2009.2009+2009}{2009.2009+2010}\)
=2008/2009-2009/2008+1/2009+2007/2008
=(2008/2009+1/2009)-(2009/2008-2007/2008)
=1-1/1004
=1003/1004
\(\frac{2008}{2009}-\frac{2009}{2008}+\frac{1}{2009}+\frac{2007}{2008}\)
=\(\left(\frac{2008}{2009}+\frac{1}{2009}\right)-\left(\frac{2009}{2008}-\frac{2007}{2008}\right)\)
=\(1-\frac{2}{2008}\)
=\(1-\frac{1}{1004}\)
=\(\frac{1003}{1004}\)
Chúc bn làm tốt nha!!!
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2008\cdot\frac{47}{3}\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=2009-\frac{94376}{3}\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right)\) \(=-\frac{88349}{3}\)
\(\frac{41}{9}+x\) \(=-\frac{88349}{3}+\frac{133}{18}\)
\(\frac{41}{9}+x\) \(=-\frac{529961}{18}\)
\(x=-\frac{529961}{18}-\frac{41}{9}\)
\(x=-\frac{176681}{6}\)
\(\frac{2006}{2007}< \frac{2007}{2007}=1\)
\(\frac{2007}{2008}< \frac{2008}{2008}=1\)
\(\frac{2008}{2009}< \frac{2009}{2009}=1\)
\(\Rightarrow a=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}< 1+1+1=3\)
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}\)
\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(A=3-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)< 3\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right):15\frac{2}{3}=2008\)
\(2009-\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2008\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=2009-2008\)
\(\left(\frac{41}{9}+x-\frac{133}{18}\right):\frac{47}{3}=1\)
\(\frac{41}{9}+x-\frac{133}{18}=1\cdot\frac{47}{3}\)
\(\frac{41}{9}+x-\frac{133}{18}=\frac{47}{3}\)
\(\frac{41}{9}+x=\frac{133}{18}+\frac{47}{3}\)
\(\frac{41}{9}+x=\frac{415}{18}\)
\(x=\frac{415}{18}-\frac{41}{9}\)
\(x=\frac{37}{2}\)
\(2009-\left(4\frac{5}{9}+x-7\frac{7}{18}\right)\div15\frac{2}{3}=2008\)
\(\left(4\frac{5}{9}+x-7\frac{7}{18}\right)\div15\frac{2}{3}=2009-2008\)
\(\left(-3\frac{3}{18}+x\right)\div15\frac{2}{3}=1\)
\(-3\frac{3}{18}+x=15\frac{2}{3}\)
\(x=15\frac{2}{3}+3\frac{3}{18}\)
\(x=15\frac{12}{18}+3\frac{3}{18}\)
\(x=18\frac{15}{18}\)
exactly 100%
29/28 nhé