Question

[TFBoys]

giải phương trình 2x⁴-7x³+9x²-7x+2=0

Answer 1

[hattori heiji]

vì x=0 không là nghiệm của pt => chia cả 2 vế cho x²≠0

2x²-7x+9-\(\dfrac{7}{x}\)+\(\dfrac{2}{x^2}\)=0

<=>\(\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+9=0\)

<=>\(2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+9=0\)

đặt \(x+\dfrac{1}{x}\)=y =>\(x^2+\dfrac{1}{x^2}=y^2-2\) ta đc

2(y²-2)-7y+9=0

<=> 2y²-4-7y+9=0

<=>2y²-7y+5=0

<=> 2y²-2y-5y+5=0

<=> (2y²-2y)-(5y-5)=0

<=> 2y(y-1)-5(y-1)=0

<=>(y-1)(2y-5)=0

<=>\(\left\{{}\begin{matrix}y-1=0\\2y-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\y=\dfrac{5}{2}\end{matrix}\right.\)

Với y=1 ta có

\(x+\dfrac{1}{x}=1\) =>x²-x+1=0 (vô nghiệm)

Với y=5/2

\(x+\dfrac{1}{x}=\dfrac{5}{2}\) => x=2 và x=\(\dfrac{1}{2}\)

vậy pt có S=\(\left\{2;\dfrac{1}{2}\right\}\)

Answer 2

[Aki Tsuki]

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

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

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

\(\Leftrightarrow2x^2\left(2x^2-2x+2\right)-\dfrac{1}{2}x\left(2x^2-2x+2\right)-2x\left(2x^2-2x+2\right)+\left(2x^2-2x+2\right)=0\)

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

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

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

Vì: \(2x^2-2x+2=\left(\sqrt{2}x-\dfrac{\sqrt{2}}{2}\right)^2+\dfrac{3}{2}>0\forall x\)

Nên: \(\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy..................

p/s: 1 cách khác :))