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\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\...\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\)
\(\Rightarrow11x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\)
=\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}\)
\(=10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)
\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{11-10}{10.11}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=10x+A=10x+\frac{10}{11}=11x\)
\(\Rightarrow\frac{10}{11}=11x-10x\)
\(\Rightarrow x=\frac{10}{11}\)
đk: \(\begin{cases}x+2\ne0\\4-x>0\\6+x>0\end{cases}\)
ta có \(3\log_{\frac{1}{4}}\left(x+2\right)-3=3\log_{\frac{1}{4}}\left(4-x\right)+3\log_{\frac{1}{4}}\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right)-\log_{\frac{1}{4}}\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\log_{\frac{1}{4}}\left(x+2\right).\frac{1}{4}=\log_{\frac{1}{4}}\left(4-x\right)\left(6+x\right)\) suy ra \(\frac{x+2}{4}=\left(4-x\right)\left(6+x\right)\)
giải pt tìm ra x
đối chiếu với đk của bài ta suy ra đc nghiệm của pt
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2010}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2009}{2010}\)
\(=\frac{1.2.3.4.5....2008.2009}{2.3.4....2009.2010}\)
\(=\frac{1}{2010}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2010}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right).\left(\frac{3}{3}-\frac{1}{3}\right).\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{2010}{2010}-\frac{1}{2010}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2009}{2010}=\frac{1.2.3....2009}{2.3.4....2010}=\frac{1}{2010}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)..................\left(1-\frac{1}{20}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.............\frac{19}{20}\)
=\(\frac{1.2.3..............19}{2.3.4..............20}\)
=\(\frac{1}{20}\)
Ta đã biết: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
Ta có: \(A=1+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+...+\frac{1}{20}.\left(\frac{20.21}{2}\right)\)
\(A=1+\frac{3}{2}+\frac{4}{2}+....+\frac{21}{2}\)
\(A=\frac{1}{2}.\left(2+3+....+21\right)\)
Tổng trong ngoặc có:21-2+2=20 (số hạng)
\(=>A=\frac{1}{2}.\left(\frac{\left(21+2\right).20}{2}\right)=\frac{1}{2}.230=115\)
Vậy..........
\(\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(-\frac{1}{3}-1\right)\)
\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\left(-\frac{4}{3}\right)\)
\(=\left[\frac{2}{3}+1+1\right]:\left(-\frac{4}{3}\right)\)
\(=\frac{8}{3}.\frac{-3}{4}\)
\(=-2\)
help me T×m mét sè cã ba ch÷ sè, biÕt r»ng sè ®ã chia hÕt cho 18 vµ c¸c ch÷ sè cña nã tØ lÖ víi ba sè 1, 2 vµ 3.