Giải PT:
|x-2016|=2016x
Những câu hỏi liên quan
Giải PT:
| x - 2016 | = 2016x
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 \(x=\frac{2016}{2017}\)
giải pt:
\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)
Khó qúa, Ai giải giùm với
??
\(2015\sqrt{2015x-2014}+\sqrt{2016x-2015}=2016\)
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: \(\frac{2015^2}{\sqrt{2015x-2014}+1}+\frac{2016}{\sqrt{2016x-2015}+1}>0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
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Tính A = x2016 - 2016.x2015 + 2016.x2014 - 2016.x2013 + ... + 2016x2 - 2016x +2016 tại x = 2015
x=2015
=> x+1=2016
=> A=x2016-(x+1).x2015+(x+1).x2014-(x+1).x2013+...+(x+1)x2-(x+1)x+2016
=x2016-x2016-x2015+x2015+x2014-x2014-x2013+...+x3+x2-x2-x+2016
=-x+2016
=-2015+2016
=1
Vậy A=1.
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\(x^{2017}-2016x^{2016}-2016x^{2015}-...-2016x^3-2016x^2-2016x\)
Cho đa thức :
A(x) = 2016- 2016x + 2016 x2 - 2016x3 + ... - 2016x2015 + x2016
Tính A(2015)?
A(x)=x^21-2016x^20+2016x^19-2016x^18+...+2016x^3-2016^2+2016x-1
Tính gá trị đa thức A(x) tại x=2015
Tính giá trị của đa thức:
P(x) = x^{2017}-2016x^{2016}-2016x^{2015}-...--2016x^2^-2016x+1 tại x=2017
Cho f(x)=x^2017-2016x^2016+2016x^2015-...+2016x-1. Tính f(2015)
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
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Giải phương trình: \(x^4\sqrt{x+3}=2x^4-2016x+2016\)
\(x\ge-3\)
\(x^4\left(\sqrt{x+3}-2\right)+2016\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{x^4\left(x-1\right)}{\sqrt{x+3}+2}+2016\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{x^4}{\sqrt{x+3}+2}+2016\right)=0\)
\(\Leftrightarrow x-1=0\) (do \(\dfrac{x^4}{\sqrt{x+3}+2}+2016>0\) \(\forall x\ge-3\) )
\(\Rightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
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