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

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

      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}

x^4-30x^2+31x-30=0

<=>x^4+x^2+1-31(x^2-x+1)=0

<=>(x^2-x+1)(x^2+x+1)-31(x^2-x+1)=0

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

<=>(x^2-x+1)(x^2-6x+5x-30)=0

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

Ta có:x^2-x+1=x^2-x+1/4+3/4=(x-1/2)^2+3/4>0 Với mọi x

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

<=>x+5=0<=>x=-5

     x-6=0<=>x=6

Vậy x=(5;-6)

= x^4+x^2+1-31x^2+31x-31

= (x^2+x+1)(x^2-x+1)-31(x^2-x+1)

= (x^2-x+1)(x^2+x+1-31)

= (x^2-x-1)(x^2+x-30)

 =  (x^2-x+1)(x^2+6x-5x-30)

=    (x^2-x+1)(x-5)(x+6)

vũ mạnh phi sai ở dấu = thứ  ấy là cộng 1 chớ ko phải trừ 1

x^4-30x^2+31x-30=0 
<=>(x^4 - 29x^2 + 841/4) - (x^2 - 31x + 31^2/4 ) =0 
<=> (x^2- 29/2)^2 - (x-31/2)^2=0 
(đến đây ta giải phương trình A^2-B^2=0 bằng cách đưa về pt tích (A-B)(A+B)=0 )

tick nha

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

\(\Leftrightarrow x^4+x-30\left(x^2-x+1\right)=0\)

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

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

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-5=0\\x^2-x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=5\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\left(loai\right)\end{array}\right.\)

Vậy \(S=\left\{-6;5\right\}\)