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a) \(\frac{10}{17}-\frac{5}{13}+\frac{7}{17}+\frac{-8}{13}-\frac{11}{25}\)
\(=\frac{10}{17}+\frac{-5}{13}+\frac{7}{17}+\frac{-8}{13}+\frac{-11}{25}\)
\(=\left(\frac{10}{17}+\frac{7}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)+\frac{-11}{25}\)
\(=1+\left(-1\right)+\frac{-11}{25}\)
\(=0+\frac{-11}{25}\)
\(=\frac{-11}{25}\)
a) \(\frac{10}{17}-\frac{5}{13}+\frac{7}{17}+\frac{-8}{13}-\frac{11}{25}\)
\(=\frac{10}{17}+\frac{-5}{13}+\frac{7}{17}+\frac{-8}{13}+\frac{-11}{15}\)
\(=\left(\frac{10}{17}+\frac{7}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)+\frac{-11}{15}\)
\(=1+\left(-1\right)+\frac{-11}{15}\)
\(=0+\frac{-11}{15}\)
\(=\frac{-11}{15}\)
b) 1 - 2 + 3 - 4 + ...... + 2011 - 2012 ( có 2012 số )
= ( 1 - 2 ) + ( 3 - 4 ) + ....... + ( 2011 - 2012 ) ( có 1006 nhóm )
= ( - 1 ) + ( - 1 ) + ........ + ( - 1 ) ( có 1006 số )
= ( - 1 ) . 1006
= - 1006
\(b)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Vậy \(\frac{2009^{2009}+1}{2009^{2010}+1}>\frac{2009^{1010}-2}{2009^{2011}-2}\)
Chúc bạn học tốt ~
Àk mình còn thiếu một điều kiện nữa xin lỗi nhé :
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Bạn thêm vào nhé
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
MS=\(\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{2}{2011}\right)+\left(1+\frac{1}{2012}\right)\)
=\(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
=> \(x.\frac{1}{2013}=1\)
=>x=2013
\(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}=\frac{16}{20}-\frac{15}{20}-\frac{1}{20}=0\)
Giá trị biểu thức đã cho là 0
\(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{2}\right)\)
\(=\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
\(=0\)
Vế phải lấy MSC là 20 rồi quy đồng lên là ok nhé bạn :))
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x\) = \(\frac{1}{10}+\frac{1}{2}\)
\(\frac{2}{3}x\) = \(\frac{3}{5}\)
\(x\) = \(\frac{3}{5}:\frac{2}{3}\)
\(x\) = \(\frac{9}{10}\)
\(\frac{1}{3}\)---- 25 % là???
\(Q=\left(\frac{1}{3}-25\%-\frac{1}{2}\right)\cdot\left(\frac{-789}{2011}+\frac{5}{2012}-17\right)\)
\(Q=\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\cdot\left(\frac{-789}{2011}+\frac{5}{2012}-\frac{17}{1}\right)\)
\(Q=\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\cdot\left(\frac{-1587468}{4046132}+\frac{10055}{4046132}-\frac{68784244}{4046132}\right)\)
\(Q=\frac{-5}{12}\cdot\frac{-70361657}{4046132}=\frac{351808285}{48553584}\)