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29 tháng 1 2018

       \(x^4+2015x^2+2014x+2015=0\)

\(\Leftrightarrow\)\(\left(x^4+x^2+1\right)+\left(2014x^2+2014x+2014\right)=0\)

\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2014\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+2015\right)=0\)

Ta có:   \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)

           \(\left(x-\frac{1}{2}\right)^2+2014\frac{3}{4}>0\)

Vậy  pt  vô nghiệm

29 tháng 1 2018

ai làm hộ mk với 

tks nhiều

ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2014};-\dfrac{2}{2015};-\dfrac{3}{2016};-\dfrac{4}{2017}\right\}\)

Ta có: \(\dfrac{1}{2014x+1}-\dfrac{1}{2015x+2}=\dfrac{1}{2016x+3}-\dfrac{1}{2017x+4}\)

\(\Leftrightarrow\dfrac{2015x+2-2014x-1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{2017x+4-2016x-3}{\left(2016x+3\right)\left(2017x+4\right)}\)

\(\Leftrightarrow\dfrac{x+1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{x+1}{\left(2016x+3\right)\left(2017x+4\right)}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}-\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\dfrac{1}{\left(2014x+1\right)\left(2015x+2\right)}=\dfrac{1}{\left(2016x+3\right)\left(2017x+4\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\4058210x^2+6043x+2=4066272x^2+14115x+12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8072x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x^2+8062x+10x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\8062x\left(x+1\right)+10\left(x+1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\\left(x+1\right)\left(8062x+10\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=0\\8062x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-1\\8062x=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(nhận\right)\\x=\dfrac{-5}{4031}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{-5}{4031}\right\}\)

21 tháng 2 2021

thanks

 

13 tháng 8 2015

 

 x4+2015x2+2014x+2015

=x4-x+2015x2+2015x+2015

=x.(x3-1)+2015.(x2+x+1)

=x.(x-1)(x2+x+1)+2015.(x2+x+1)

=(x2+x+1)(x2-x+2015)

13 tháng 8 2015

\(x^4+2015x^2+2014x+2015=\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(2015x^2+2015x+2015\right)\)

\(=x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)

23 tháng 8 2021

\(x^4+2015x^2+2014x+2015.\)

=\(\left(x^4-x\right)+2015x^2+2015x+2015\)

=\(x\left(x^3-1\right)+2015\left(x^2+x+1\right)\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)\)

\(\left(x^2+x+1\right)\left(x^2-x-2015\right)\)

k cho mik

trả lời

xx^4+2015x^2+2014x+2015=x^4+2015x^2+2015x-x+2015=x\left(x^3-1\right)+2015\left(X^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)xx

13 tháng 6 2019

\(x^4+2015x^2+2014x+2015\)

\(=\left(x^4-x\right)+2015x^2+2015x+2015\)

\(=x\left(x^3-1\right)+2015\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)

1 tháng 3 2016

Câu 1 nha bạn 

x^4 + x^3 + x^2 + 2014x^2 + 2014x + 2014 + 1 - x^3

=> x^4 + x^3 + x^2 + 2014x^2 + 2014x + 2014 - x^3 - 1

=> x^2 ( x^2 + x + 1 ) + 2014 ( x^2 + x + 1 ) - ( x - 1 )( x^2 + x + 1 ) 

=> ( x^2 + x + 1 )( x^2 + 2014 - x - 1)

29 tháng 7 2016

mk nhầm đề sửa lại là phân tích đa thức thành nhân tử

29 tháng 7 2016

x4+2014x2-2014x-x+2014

=x(x3-1)+2014(x2-x-1)

=x(x-1)(x2-x-1)+2014(x2-x-1)

=(x2-x-1)(x2-x+2014)

 

30 tháng 11 2017

\(x^4+2015x^2+2014x+2015\)

\(=\left(x^4-x^3+2015x^2\right)+\left(x^3-x^2+2015x\right)+\left(x^2-x+2015\right)\)

\(=\left(x^2-x+2015\right)\left(x^2+x+1\right)\)