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

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

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

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

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

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

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

Vậy \(S=\left\{-6;5\right\}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

....

Điểm âm của tớ chỉ còn 29 nữa thôi, các bạn giúp mik nha, khi nào hết âm mik sẽ ra câu hỏi và giúp lại các bạn

Mơn các bạn trc~~~~ Và cx mơn các bạn đã giúp mik trong thời gian qua =DD

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

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

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

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

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

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

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