Tìm x biết : \(x^4-30x^2+31x-30=0\)
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20 tháng 6 2019
x^4 - 30x^2 + 31x - 30 = 0
<=> x^4 + x^3 - 30x^2 - x^3 - x^2 + 30x+ x^2 + x - 30 = 0
<=> x^2(x^2 + x - 30) - x(x^2 + x - 30) + (x^2 + x - 30) = 0
<=> (x^2 - x + 1)(x^2 + x - 30) = 0
<=> (x^2 - x + 1)(x + 6)(x - 5) = 0
Mà x^2 - x + 1 = (x^2 - 2.x.1/2 + 1/4) + 3/4 = (x - 1/2)^2 + 3/4 > 0
=> x = -6 hoặc x = 5
hc tốt ~:B~
K
20 tháng 6 2019
x^4 - 30x^2 + 31x - 30 = 0
<=> x^4 + x^3 - 30x^2 - x^3 - x^2 + 30x+ x^2 + x - 30 = 0
<=> x^2(x^2 + x - 30) - x(x^2 + x - 30) + (x^2 + x - 30) = 0
<=> (x^2 - x + 1)(x^2 + x - 30) = 0
<=> (x^2 - x + 1)(x + 6)(x - 5) = 0
Mà x^2 - x + 1 = (x^2 - 2.x.1/2 + 1/4) + 3/4 = (x - 1/2)^2 + 3/4 > 0
=> x = -6 hoặc x = 5
S
4 tháng 1 2019
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)
Ta có: \(x^2-x+1=x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge0\forall x\in R\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
Vậy, \(S=\left\{-6;5\right\}\)
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
SD
1
N
10 tháng 10 2017
x^4-30x^2+31x^2-30=0
=>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+1)(x^2-5x+6x-30)=0
=>(x^2-x+1)[x(x-5)+6(x-5)]=0
=>(x^2-x+1)(x-5)(x+6)=0
vì x^2-x+1=x^2-2x.(1/2)+1/4+3/4
=(x-1/2)^2+3/4>0 với mọi x
Do đó x-5=0 hoac x+6=0
=>x=5 hoặc x=-6
NN
22 tháng 11 2017
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2-5x+6x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[\left(x^2-5x\right)+\left(6x-30\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x-5\right)+6\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x-5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x-5=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(loai\right)\\x=5\\x=-6\end{matrix}\right.\)
Vậy x=5 hoặc x=-6
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 \(S=\left\{-6;5\right\}\)
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
\(x^4-30x^2+31x-30=0\)
\(\left(x^4+x\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
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 \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Thanks bạn nha