Giải PT:
|x-2016|=2016x
K
Khách
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KN
14 tháng 2 2020
We have two cases:
+) If \(x\ge2016\)then \(x-2016\ge0\Rightarrow\left|x-2016\right|=x-2016\)
Equation becomes: \(x-2016=2016x\)
\(\Leftrightarrow2015x=-2016\Leftrightarrow x=\frac{-2016}{2015}\)(not satisfied)
+) If \(x< 2016\)then \(x-2016< 0\Rightarrow\left|x-2016\right|=2016-x\)
Equation becomes: \(2016-x=2016x\)
\(\Leftrightarrow2017x=2016\Leftrightarrow x=\frac{2016}{2017}\)(satisfied)
So \(x=\frac{2016}{2017}\)
TH
1
TN
23 tháng 9 2017
\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)
ĐK:\(x\ge\frac{2015}{2016}\)
\(\Leftrightarrow2015\left(\sqrt{2015x-2014}-1\right)+\sqrt{2016x-2015}-1=0\)
\(\Leftrightarrow2015\frac{2015x-2014-1}{\sqrt{2015x-2014}+1}+\frac{2016x-2015-1}{\sqrt{2016x-2015}+1}=0\)
\(\Leftrightarrow2015\frac{2015x-2015}{\sqrt{2015x-2014}+1}+\frac{2016x-2016}{\sqrt{2016x-2015}+1}=0\)
\(\Leftrightarrow2015\frac{2015\left(x-1\right)}{\sqrt{2015x-2014}+1}+\frac{2016\left(x-1\right)}{\sqrt{2016x-2015}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}\right)=0\)
Dễ thấy: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
NV
0
LT
0
LH
Tính giá trị của đa thức:
P(x) = x^{2017}-2016x^{2016}-2016x^{2015}-...--2016x^2^-2016x+1 tại x=2017
0
US
1
LP
4 tháng 1 2017
Theo đề bài ta có
\(f\left(x\right)=x^{2017}-2016.x^{2016}+2016.x^{2015}-...+2016.x-1\)
Với \(f\left(2015\right)\)thì \(x=2015,x+1=2016\)
\(\Rightarrow f\left(x\right)=x^{2017}-\left(x+1\right).x^{2016}+\left(x+1\right).x^{2015}-...+\left(x+1\right).x-1\)
\(\Rightarrow f\left(x\right)=x^{2017}-x^{2017}-x^{2016}+x^{2016}+x^{2015}-...+x^2+x-1\)
\(\Rightarrow f\left(x\right)=x-1\)
\(\Rightarrow f\left(2015\right)=2015-1=2014\)
Vậy f(2015)=2014
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2018
\(x\ge-3\)
\(x^4\left(\sqrt{x+3}-2\right)+2016\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{x^4\left(x-1\right)}{\sqrt{x+3}+2}+2016\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x^4}{\sqrt{x+3}+2}+2016\right)=0\)
\(\Leftrightarrow x-1=0\) (do \(\dfrac{x^4}{\sqrt{x+3}+2}+2016>0\) \(\forall x\ge-3\) )
\(\Rightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
ĐK: x>0.
Pt\(\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2015}{2016}\left(l\right)\\x=\frac{2016}{2017}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{2016}{2017}\right\}\)
Ta có: \(\left|x-2016\right|=2016x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2016=2016x\\x-2016=-2016x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2016-2016x=0\\x-2016+2016x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2015x=2016\\2017x=2016\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2016}{2015}\\x=\frac{2016}{2017}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2016}{2015};\frac{2016}{2017}\right\}\)