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x^4-30x^2+31x-30
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14 tháng 12 2016
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-5\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-5=0\\x^2-x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=5\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\left(loai\right)\end{array}\right.\)
Vậy \(S=\left\{-6;5\right\}\)
25 tháng 2 2017
\(=\left(x^4+x\right)-30x^2+30x-30\)
\(=x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(=x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]\)
\(=\left(x^2-x+1\right)\left(x^2+x-30\right)\)
3 tháng 2 2020
x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30=0
(x-5)(x^3+5x^2-5x+6)=0
(x-5)(x^3+6x^2-x^2-6x+x+6)=0
(x-5)(x+6)(x^2-x+1)=0
Suy ra x-5=0 hay x+6=0 hay x^2-x+1=0
Suy ra x=5 hay x=-6 hay x^2+2x.1/2+1/4+3/4=0
Suy ra x=5 hay x=-6 hay (x+1/2)^2=3/4=0 (vô lý)
Vậy x=5 hay x=-6
5 tháng 6 2018
\(x^4-30x^2+31x-30\)
\(=x^4+x-30x^2+30x-30\)
\(=x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(=x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(=\left(x^2+x\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x-30\right)\)
6 tháng 6 2018
\(x^4-30x^2+31x-30\)
\(=x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30\)
\(=x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^3+5x^2-5x+6\right)\)
\(=\left(x-5\right)\left(x^3+6x^2-x^2-6x+x+6\right)\)
\(=\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]\)
\(=\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)\)
21 tháng 2 2018
pt <=> (x^4+x)-(30x^2-30x+30) = 0
<=> x.(x^3+1)-30.(x^2-x+1) = 0
<=> x.(x+1).(x^2-x+1)-30.(x^2-x+1) = 0
<=> (x^2-x+1).(x^2+x-30) = 0
<=> x^2+x-30 = 0 ( vì x^2-x+1 > 0 )
<=> (x^2-5x)+(6x-30) = 0
<=> (x-5).(x+6) = 0
<=> x-5=0 hoặc x+6=0
<=> x=5 hoặc x=-6
Vậy ..............
Tk mk nha
14 tháng 6 2017
x4-30x2+31x-30=0
x4+x) -30x2+30x-30=0
x{x3+1} -30{ x2-x+1}=0
x{x+1}{x2-x+1}-30{x2-x+1}=0
{x2-x+1}{x2+x-30}=0
x2+x-30=0 {vi x2-x+1>0}
x2+x-30x-30=0
{x+1}{x-30}=0
- x=-1
- x=30
9 tháng 12 2018
\(x^4-30x^2+31x-30=0\)
\(\left(x^4+x\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
Ta có: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2+x-30=0\left(x^2-x+1\ne0\right)\)
\(\left(x^2-5x\right)+\left(6x-30\right)=0\)
\(x\left(x-5\right)+6\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
x4-30x2+31x-30
=(x4+x)-(30x2-30x+30)
=x(x3+1)-30(x2-x+1)
=x(x+1)(x2-x+1)-30(x2-x+1)
=(x(x+1)-30)(x2-x+1)