Question

[Lê Phúc Thuận]

Giai phuong trinh :

2x^4-7x^3+9x^2-7x+2=0

Answer 1

[Đinh Đức Hùng]

\(2x^4-7x^3+9x^2-7x+2=0\)

\(\Leftrightarrow2x^4-x^3-6x^3+3x^2+6x^2-3x-4x+2=0\)

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

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

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

Ta dễ thấy \(x^3-3x^2+3x-2>0\forall x\) nên để PT (1) có nghiệm \(\Leftrightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)

Vậy nghiệp phương trình trên là \(S=\left\{\frac{1}{2}\right\}\)

Answer 2

[Đinh Đức Hùng]

Sủa chút : \(\left(2x-1\right)\left(x^3-3x^2+3x-2\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}\)