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Ta có \(x=\dfrac{2016}{x\times\left(x+1\right)\times\left(x+2\right)\times........\times\left(x+2016\right)}\)
\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times...........\times2016}\)
Vì x > 0=> \(\left(x+1\right)\times\left(x+2\right)\times...\times\left(x+2016\right)>1\times2\times...\times2016\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{1}{1\times2\times..........\times2016}\)\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times.......\times\left(x+2016\right)}< \dfrac{2016}{1\times2\times......\times2016}\)
\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)
Ta có \(x=\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times....\times\left(x+2016\right)}\)
\(\dfrac{1}{2015!}=\dfrac{2016}{2016!}=\dfrac{2016}{1\times2\times.....\times2016}\)
Vì x>0=>(x+1)×(x+2)×.............×(x+2016) >\(1\times2\times.....\times2016\)
\(\Rightarrow\dfrac{1}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{1}{1\times2\times......\times2016}\)
\(\Rightarrow\dfrac{2016}{\left(x+1\right)\times\left(x+2\right)\times......\times\left(x+2016\right)}>\dfrac{2016}{1\times2\times......\times2016}\)
\(\Leftrightarrow x< \dfrac{1}{2015!}\)(đpcm)
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
\(\frac{\left|x\right|+2015}{2016}\) . Có: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2015\ge2015\Rightarrow\frac{\left|x\right|+2015}{2016}\ge\frac{2015}{2016}\)
Dấu = xảy ra khi \(x+2015=0\Rightarrow x=0\)
Vậy \(Min\frac{\left|x\right|+2015}{2016}=\frac{2015}{2016}\) tại \(x=0\)
\(\frac{\left|x\right|+1996}{-1997}\) có \(\left|x\right|\ge0\Rightarrow\left|x\right|+1996\ge1996\Rightarrow\frac{\left|x\right|+1996}{-1997}\le-\frac{1996}{1997}\)
Dấu = xảy ra khi \(\left|x\right|+1996=1996\Rightarrow x=0\)
Vậy \(Max\frac{\left|x\right|+1996}{-1997}=\frac{1996}{-1997}\) tại \(x=0\)
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