Cho B= \(15+\frac{17}{10+\frac{4}{2009}}\). Viết B ở dạng [ a0;a1;a2;......;an]
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a) \(\frac{2}{3} = \frac{4}{6} = \frac{1}{6} + \frac{3}{6} = \frac{1}{6} + \frac{1}{2}\)
b) \(\frac{8}{{15}} = \frac{5}{{15}} + \frac{3}{{15}} = \frac{1}{5} + \frac{1}{3}\)
c) \(\frac{7}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8}\)
d) \(\frac{{17}}{{18}} = \frac{9}{{18}} + \frac{6}{{18}} + \frac{2}{{18}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{9}\).
cu lay phep tinh nay tru phep tinh kia hk ra thi nt hoi mink
a)
\(\begin{array}{l}\frac{4}{{15}} - \left( {2,9 - \frac{{11}}{{15}}} \right)\\ = \frac{4}{{15}} - 2,9 + \frac{{11}}{{15}}\\ = \left( {\frac{4}{{15}} + \frac{{11}}{{15}}} \right) - 2,9\\=\frac{15}{15}-2,9 \\= 1 - 2,9 = - 1,9\end{array}\)
b)
\(\begin{array}{l}( - 36,75) + \left( {\frac{{37}}{{10}} - 63,25} \right) - ( - 6,3)\\ = ( - 36,75) + 3,7 - 63,25 + 6,3\\ = \left( { - 36,75 - 63,25} \right) + \left( {3,7 + 6,3} \right)\\ = - 100 + 10 = - 90\end{array}\)
c)
\(\begin{array}{l}6,5 + \left( { - \frac{{10}}{{17}}} \right) - \left( { - \frac{7}{2}} \right) - \frac{7}{{17}}\\ = \frac{{65}}{{10}} - \frac{{10}}{{17}} + \frac{7}{2} - \frac{7}{{17}}\\ = \left( {\frac{{65}}{{10}} + \frac{7}{2}} \right) - \left( {\frac{{10}}{{17}} + \frac{7}{{17}}} \right)\\ = \left( {\frac{{65}}{{10}} + \frac{{35}}{{10}}} \right) - \frac{17}{17}\\ = \frac{100}{10}-1\\=10 - 1 = 9\end{array}\)
d)
\(\begin{array}{l}( - 39,1) \cdot \frac{{13}}{{25}} - 60,9 \cdot \frac{{13}}{{25}}\\ = \frac{{13}}{{25}}.\left( { - 39,1 - 60,9} \right)\\ = \frac{{13}}{{25}}.\left( { - 100} \right)\\ = - 52\end{array}\).
Ta có :
\(17A=\frac{17^{2009}+17}{17^{2009}+1}=\frac{17^{1009}+1+16}{17^{2009}+1}=\frac{17^{2009}+1}{17^{2009}+1}+\frac{16}{17^{2009}+1}=1+\frac{16}{17^{2009}+1}\)
\(17B=\frac{17^{2010}+17}{17^{2010}+1}=\frac{17^{2010}+1+16}{17^{2010}+1}=\frac{17^{2010}+1}{17^{2010}+1}+\frac{16}{17^{2010}+1}=1+\frac{16}{17^{2010}+1}\)
Vì \(\frac{16}{17^{2009}+1}>\frac{16}{17^{2010}+1}\) nên \(17A>17B\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Đặt \(A=\frac{15+\frac{15}{7}-\frac{15}{11}+\frac{15}{2009}-\frac{15}{13}}{\frac{4}{2009}-\frac{4}{13}+\frac{4}{7}-\frac{4}{11}+4}\)
\(=\frac{15\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}{4\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}\)
\(=\frac{15}{4}\)
Đặt \(B=\frac{5\cdot2010-1996}{14+4\cdot2010}\)
\(=\frac{5\left(1996+4\right)-1996}{14+4\cdot2010}\)
\(=\frac{5\cdot1996+20-1996}{14+4\left(1996+4\right)}\)
\(=\frac{4\cdot1996+20}{4\cdot1996+30}\)
\(\Rightarrow A\cdot B=\frac{4\cdot1996+20}{4\cdot1996+30}\cdot\frac{15}{4}=\frac{15\cdot4\left(1996+5\right)}{4\left(4\cdot1996+30\right)}=\frac{15\left(1996+5\right)}{4\cdot1996+30}=\frac{30015}{8004}\)
mặc dầu ko khoa học lắm nhưng mình thấy cũng được đấy
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
tìm n N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên
giúp mik với sắp thi r
\(A=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(-\frac{20}{31}-\frac{11}{31}\right)+-\frac{4}{9}\)
\(A=1+\left(-1\right)+\frac{-4}{9}\)
\(A=0+\frac{-4}{9}=\frac{-4}{9}\)
\(B=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{3}\)
\(B=-1+1+\frac{-2}{3}\)
\(B=\frac{-2}{3}\)