Question

[Trần Khánh Đan]

Giải phương trình sau:

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

Answer 1

[Nguyễn Huy Tú]

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

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

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

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

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

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

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

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

Vì \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)

Vậy x = -6 hoặc x = 5

Answer 2

[An Trần]

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

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

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

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+1=0\\x^2+x-30=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vl\right)\\\left(x-5\right)\left(x+6\right)=0\end{matrix}\right.\)

=> x = 5 hoặc x = -6.

p/s: ***** = vô lý :V