Tìm x biết: \(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\)
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\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)
\(=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}.....\frac{2006}{2007}.\frac{2007}{2008}\)
\(=\frac{9.10.11.....2006.2007}{10.11.12.....2007.2008}\)
\(=\frac{9}{2008}\)
\(Ta\) \(có:\)
\(A=\frac{9}{2008}\)
\(B=\frac{1}{2000}\)
\(\frac{9}{2008}=\frac{9.250}{2008.250}=\frac{2250}{502000}\)
\(\frac{1}{2000}=\frac{1.251}{2000.251}=\frac{251}{502000}\)
Vì \(\frac{2250}{502000}>\frac{251}{502000}\Rightarrow A>B\)
\(A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{2007}\right)\left(1-\frac{1}{2008}\right)\)
\(A=\frac{9}{10}.\frac{10}{11}.\frac{11}{12}....\frac{2006}{2007}.\frac{2007}{2008}\)
\(A=\frac{9.10.11....2006.2007}{10.11.12...2007.2008}\)
\(A=\frac{9}{2008}\)
Vì \(\frac{9}{2008}
=> 1/2.2/3.3/4...x/(x+1)=1/2009
=> 1/(x+1)=1/2009
=> x+1=2009
=> x=2008
Mik nhanh nhất mik nha
=> 1/2 . 2/3 . 3/4 .... x/x+1 = 1/2009
<=> (1.2.3...x) / [2.3...x.(x+1)] = 1/2009
<=> 1/ (x+1) = 1/2009
=> x+1=2009 => x=2008
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2009}\right)\)
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2008}{2009}\)
=\(\frac{1}{2009}\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\)
\(=\frac{1}{6}\)
\(\left(1-\frac{1}{2}\right)\)x \(\left(1-\frac{1}{3}\right)\)x \(\left(1-\frac{1}{4}\right)\)x \(\left(1-\frac{1}{5}\right)\)x \(\left(1-\frac{1}{6}\right)\)
= \(\frac{1}{2}\)x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\)
= \(\frac{1x2x3x4x5}{2x3x4x5x6}\)
= \(\frac{1}{6}\)
k mình nha
Chúc bạn học giỏi
Mình cảm ơn bạn nhiều
(1-1/2)(1-1/3)(1-1/4)...(1-1/2009)
=1/2*2/3*3/4*...*2008/2009
=\(\frac{1\cdot2\cdot3\cdot...\cdot2008}{2\cdot3\cdot4\cdot...\cdot2009}\)
=1/2009
\(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1\le0\\\frac{2}{3}x+\frac{1}{5}\le0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x\le1\\\frac{2}{3}x\le-\frac{1}{5}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\le\frac{1}{3}\\x\le-\frac{3}{10}\end{array}\right.\)
\(\Rightarrow x\le\frac{1}{3}\left(tm\right)\)
Vậy để \(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\) thì \(x\le\frac{1}{3}\)