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HB
0
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)
8 tháng 5 2018
a)Đặt \(A=\dfrac{1}{8}x^3-\dfrac{3}{4}x^2+\dfrac{3}{2}x-1\)
\(A=\dfrac{1}{8}\left(x^3-6x^2+12x-8\right)\)
\(A=\dfrac{1}{8}\left(x-2\right)^3\)
8 tháng 5 2018
b,\(x^4+2015x^2+2014x+2015=x^4+2015x^2+2015x-x+2015=x\left(x^3-1\right)+2015\left(X^2+x+1\right)=x\left(x-1\right)\left(x^2+x+1\right)+2015\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^2-x+2015\right)\)
PT
0
TQ
2
21 tháng 2 2021
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: \(S=\left\{-1;\dfrac{-5}{4031}\right\}\)
TN
0
PT
25 tháng 10 2016
1.x2-9
= (x-3)(x+3)
2. -2x2+2x+12
= -2x2+6x-4x+12
= -2x(x+2)+6(x+2)
= (x+2)(-2x+6)
4. -2x2+2x+24
= -2x2+8x-6x+24
= -2x(x+3)+8(x+3)
= (x+3)(-2x+8)
6. x2-5x+4
= x2-4x-x+4
= x(x-1) -4(x-1)
= (x-1)(x-4)
8. x2-7x+6
= x2-6x-x+6
= x(x-1)-6(x-1)
= (x-1)(x-6)
9. x2+5x+4
= x2+4x+x+4
= x(x+1)+4(x+1)
=(x+1)(x+4)
10. x2+7x+6
= x2 +x+6x+6
= x(x+1)+6(x+1)
= (x+6)(x+1)
K nhé
11 tháng 1 2017
Theo bài ra , ta có :
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy \(x=-2016\)
Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
Chúc bạn học tốt =))
11 tháng 1 2017
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
\(x^4+2015x^2+2014x+2015=0\)
\(\Leftrightarrow\)\(\left(x^4+x^2+1\right)+\left(2014x^2+2014x+2014\right)=0\)
\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+1\right)+2014\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(\left(x^2+x+1\right)\left(x^2-x+2015\right)=0\)
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
ai làm hộ mk với
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