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A=\(\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\cdot\cdot\dfrac{-2015}{2016}\)
=\(-\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\cdot\cdot\dfrac{2015}{2016}\)
=\(\dfrac{-1}{2016}>\dfrac{-1}{2015}\)
Vậy\(A>\dfrac{-1}{2015}\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)
\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)
=>x=13/12
b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)
\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)
\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)
d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)
\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)
e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)
=>x+2020=0
hay x=-2020
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
a/ \(\dfrac{1}{3}-\dfrac{2}{5}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{-1}{15}+3x=\dfrac{3}{4}\)
\(\Leftrightarrow3x=\dfrac{49}{60}\)
\(\Leftrightarrow x=\dfrac{49}{180}\)
Vậy....
b/ \(\dfrac{3}{2}-1+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow\dfrac{1}{2}+4x=\dfrac{2}{3}-7x\)
\(\Leftrightarrow4x+7x=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\Leftrightarrow11x=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{66}\)
Vậy....
c/ \(2\left(\dfrac{3}{4}-5x\right)=\dfrac{4}{5}-3x\)
\(\Leftrightarrow\dfrac{3}{2}-10x=\dfrac{4}{5}-3x\)
\(\Leftrightarrow-10x+3x=\dfrac{4}{5}-\dfrac{3}{2}\)
\(\Leftrightarrow-7x=-\dfrac{7}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}\)
Vậy .....
d/ \(4\left(\dfrac{1}{2}-x\right)-5\left(x-\dfrac{3}{10}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow2-4x-5x-\dfrac{3}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow2+\left(-4x\right)+\left(-5x\right)+\left(\dfrac{-3}{2}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow-9x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\Leftrightarrow-9x=\dfrac{5}{4}\)
\(\Leftrightarrow x=-\dfrac{5}{36}\)
a: =>4x-6-9=5-3x-3
=>4x-15=-3x+2
=>7x=17
hay x=17/7
b: \(\Leftrightarrow\dfrac{2}{3x}-\dfrac{1}{4}=\dfrac{4}{5}-\dfrac{7}{x}+2\)
=>2/3x+21/3x=4/5+2+1/4=61/20
=>23/3x=61/20
=>3x=23:61/20=460/61
hay x=460/183
a/ \(\dfrac{x+1}{2}=\dfrac{2x+3}{5}\)
\(\Leftrightarrow5\left(x+1\right)=2\left(2x+3\right)\)
\(\Leftrightarrow5x+5=4x+6\)
\(\Leftrightarrow5x-4x=6-5\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ...
b/ \(\left|x-1\right|+3\left|y+1\right|+\left|z+2\right|=0\)
Mà với \(\forall x;y;z\) ta có :
\(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\3\left|y+1\right|\ge0\\\left|z+2\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\3\left|y+1\right|=0\\\left|z+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+1=0\\z+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=-2\end{matrix}\right.\)
Vậy ...
c/ \(\dfrac{x-2}{4}=\dfrac{5-3x}{4}\)
\(\Leftrightarrow x-2=5-3x\)
\(\Rightarrow x+3x=5+2\)
\(\Leftrightarrow4x=7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
Vậy ......
d/ \(\dfrac{x+2}{4}=\dfrac{4}{x+2}\)
\(\Leftrightarrow\left(x+2\right)\left(x+2\right)=16\)
\(\Leftrightarrow\left(x+2\right)^2=4^2=\left(-4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy ...
e/ \(\dfrac{x-1}{5}=\dfrac{-20}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-100\)
\(\Leftrightarrow\left(x-1\right)^2=-100\)
Lại có : \(\left(x-1\right)^2\ge0\)
\(\Leftrightarrow\) k tồn tại x
b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
a/ \(\left(4x-5\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy ............
b/ \(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x+2}{2015}+1\right)=\left(\dfrac{x+3}{2014}+1\right)+\left(\dfrac{x+4}{2013}+1\right)\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Leftrightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Leftrightarrow x+2017\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Mà \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
\(\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\)
Vậy ..
\(\left(4x-5\right)\left(3x+2\right)=0\)
\(\)\(\Rightarrow\left[{}\begin{matrix}4x-5=0\\3x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\dfrac{x+1}{2016}+\dfrac{x+2}{2015}=\dfrac{x+3}{2014}+\dfrac{x+4}{2013}\)
\(\Rightarrow\dfrac{x+1}{2016}+1+\dfrac{x+2}{2015}+1=\dfrac{x+3}{2014}+1+\dfrac{x+4}{2013}+1\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}=\dfrac{x+2017}{2014}+\dfrac{x+2017}{2013}\)
\(\Rightarrow\dfrac{x+2017}{2016}+\dfrac{x+2017}{2015}-\dfrac{x+2017}{2014}-\dfrac{x+2017}{2013}=0\)
\(\Rightarrow\left(x+2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\ne0\)
Nên:
\(x+2017=0\Rightarrow x=-2017\)