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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\(A>\dfrac{-1}{2015}\)
b) Để \(\frac{6}{x+1}.\frac{x-1}{3}\)là một số nguyên =>\(\frac{6.\left(x-1\right)}{\left(x+1\right).3}\)phải là một số nguyên
Ta có:
\(\frac{6.\left(x-1\right)}{\left(x+1\right).3}=\frac{2\left(x-1\right)}{x+1}=\frac{2\left(x+1\right)-3}{x+1}\)=> Để \(\frac{6}{x+1}.\frac{x-1}{3}\)là một số nguyên thì 2(x+1)-3 phải chia hết cho x+1
=> 3 phải chia hết cho x+1
=> x+1 thuộc vào Ư(3)=(1;-1;3;-3)
Ta có bảng
| x+1 | 1 | -1 | 3 | -3 |
| x | 0 | -2 | 2 | -4 |
Vậy x=0;-2;2;-4 thì thỏa mãn yêu cầu đề bài
Ta có : P = \(\left|a-\frac{1}{2014}\right|+\left|a-\frac{1}{2016}\right|\)
Thay a = \(\frac{1}{2015}\)vào biểu thức P ,ta có :
\(\left|\frac{1}{2015}-\frac{1}{2014}\right|+\left|\frac{1}{2015}-\frac{1}{2016}\right|\)
\(=\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(=\frac{1}{2014}-\frac{1}{2016}\)
\(=\frac{2016-2014}{2014.2016}=\frac{2}{4060224}=\frac{1}{2030112}\)
Vậy P = \(\frac{1}{2030112}\)
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:
\(x+2017=0\Rightarrow x=-2017\)
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
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
1)\(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
Cái này tự tính được nhé
2) \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\in Z\Leftrightarrow\dfrac{6\left(x-1\right)}{3\left(x+1\right)}\in Z\)
\(\Rightarrow6x-6⋮3x+3\)
\(\Rightarrow6x+6-12⋮3x+3\)
\(\Rightarrow12⋮3x+3\)
Ok:>
Câu 1:
Thay \(a=\dfrac{1}{2015}\) vào biểu thức \(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|\) ta được:
\(\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
\(=\left|\dfrac{2014}{4058210}-\dfrac{2015}{4058210}\right|+\left|\dfrac{2016}{4062240}-\dfrac{2015}{4062240}\right|\)
\(=\left|\dfrac{2014-2015}{4058210}\right|+\left|\dfrac{2016-2015}{4062240}\right|\)
\(=\left|-\dfrac{1}{4058210}\right|+\left|\dfrac{1}{4062240}\right|\)
\(=\dfrac{1}{4058210}+\dfrac{1}{4062240}\)
\(=\dfrac{1008}{4090695680}+\dfrac{1007}{4090695680}\)
\(=\dfrac{1008+1007}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{1}{2030112}\)
Vậy giá trị của biểu thức P tại \(a=\dfrac{1}{2015}\) là \(\dfrac{1}{2030112}\)