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

<=>(x ( x-1) - 30) ( x² - x + 1) =0

<=>(x² -x -30) ( x² - x + 1) =0

<=>( x² - x + 1) ( x² - 5x + 6x - 30) =0

<=> ( x² - x + 1) ( x(x-5) + 6 ( x-5)) =0

<=> ( x² - x + 1) (x-5) (x+6) =0

Vì ( x² - x + 1) > 0 với mọi x (bình phương thiếu)

=> (x-5) (x+6) =0

<=> x-5 = 0 hoặc x+ 6 = 0

<=> x=5 hoặc x = -6

30x2=60

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

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

\(x^4-30x^2+31x-30=0\)

\(\Leftrightarrow x^4-5x^3+5x^3-25x^2-5x+25x+6x-30=0\)

\(\Leftrightarrow\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)

\(\Leftrightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

Câu hỏi của trần thị anh thư - Toán lớp 8 - Học toán với OnlineMath

   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+1)(x2-5x+6x-30)=0
<=>(x2-x+1)[x(x-5)+6(x-5)]=0
<=>(x2-x+1)(x-5)(x+6)=0
Vì x2-x+1=x2-2x.1/2+1/4+3/4=(x-1/2)2+3/4>0 với mọi x
Do đó: <=>x-5 =0    <=> x=5
                x+6=0           x=-6
Vậy phương trình có tập nghiệm là S={5;-6}

P/S: kham khảo

\(x^4-30x^2+31x-30=0\)

<=>\(x^4-30x^2+30x+x-30=0\)

<=>\(\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)

<=>\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)

<=>\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)

<=>\(\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)

<=>\(\left(x^2-x+1\right)\left[\left(x^2+6x\right)-5\left(x+30\right)\right]=0\)

<=>\(x^2\left(-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)

<=>\(\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)

=>\(x+6=0hoặcx-5=0\) vì\(\left[x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\)

<=> x=-6 hoặc x=5

Vậy......

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)

=(x2+x)(x2−x+1)−30(x2−x+1)

=(x2−x+1)(x2+x−30)

tự làm tieeps nhé

     \(x^4-30x^2+31x-30=0\)

\(\Rightarrow x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30=0\)

\(\Rightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)

\(\Rightarrow\left(x-5\right)\left[x^3+6x^2-x^2-6x+x+6\right]=0\)

\(\Rightarrow\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)

\(\Rightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)

Mà \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)

Chúc bạn học tốt.