Question

[Phạm Thanh Trà]

Giải phương trình:

\(\sqrt{5x^3+3x^2+3x-2}=\frac{x^2}{2}+3x-\frac{1}{2}\)

Answer 1

[Nguyen Thi Trinh]

\(\Leftrightarrow5x^3+3x^2+3x-2=\left(\dfrac{x^2}{2}+3x-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+x^2\left(3x-\dfrac{1}{2}\right)+\left(3x-\dfrac{1}{2}\right)^2\)

\(\Leftrightarrow5x^3+3x^2+3x-2=\dfrac{x^4}{4}+3x^3-\dfrac{x^2}{2}+9x^2-3x+\dfrac{1}{4}\)

\(\Leftrightarrow20x^3+12x^2+12x-8=x^4+12x^3-2x^2+36x^2-12x+1\)

\(\Leftrightarrow x^4-8x^3+22x^2-24x+9=0\)

\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(15x^2-15x\right)-\left(9x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+15x-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)-\left(6x^2-6x\right)+\left(9x-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2-6x+9\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\left\{1;3\right\}\)