Đáp án:
`S={`$-\sqrt[]{34}-5$`;`$\sqrt[]{34}-5$`;`$-\sqrt[]{3}+2$`;`$\sqrt[]{3}+2$`}`
Giải thích các bước giải:
`x^4+6x³-48x²+46x-9=0`
`⇔x^4+10x³-4x³-9x²-40x²+x²+36x+10x-9=0`
`⇔(x^4+10x³-9x²)-(4x³+40x²-36x)+(x²+10x-9)=0`
`⇔x²(x²+10x-9)-4x(x²+10x-9)+(x²+10x-9)=0`
`⇔(x²+10x-9)(x²-4x+1)=0`
`⇔(x²+10x+25-34)(x²-4x+4-3)=0`
`⇔[(x²+10x+25)-34][(x²-4x+4)-3]=0`
`⇔[(x+5)²-34][(x-2)²-3]=0`
`⇔[(x+5)²-(`$\sqrt[]{34}$ `)^2][(x-2)^2-(`$\sqrt[]{3}$ `)^2]=0`
`⇔(x+5+`$\sqrt[]{34}$ `)(x+5-`$\sqrt[]{34}$ `)(x-2+`$\sqrt[]{3}$ `)(x-2-`$\sqrt[]{3}$ `)=0`
`⇔`\(\left[ \begin{array}{l}x+5+\sqrt[]{34}=0\\x+5-\sqrt[]{34}=0\\x-2+\sqrt[]{3}=0\\x-2-\sqrt[]{3}=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x+5=-\sqrt[]{34}\\x+5=\sqrt[]{34}\\x-2=-\sqrt[]{3}\\x-2=\sqrt[]{3}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\sqrt[]{34}-5\\x=\sqrt[]{34}-5\\x=-\sqrt[]{3}+2\\x=\sqrt[]{3}+2\end{array} \right.\)
Vậy `S={`$-\sqrt[]{34}-5$`;`$\sqrt[]{34}-5$`;`$-\sqrt[]{3}+2$`;`$\sqrt[]{3}+2$`}`
