Đáp án:
`d)S={-2;2;-3;3}`
`e)S={-`$\sqrt[]{3}$`;`$\sqrt[]{3}$`;-2;2}`
`f)S={-`$\sqrt[]{2}$`;`$\sqrt[]{2}$`}`
Giải thích các bước giải:
`d)x^4-13x²+36=0`
`⇔x^4-4x²-9x²+36=0`
`⇔x²(x²-4)-9(x²-4)=0`
`⇔(x²-4)(x²-9)=0`
`⇔(x²-2²)(x²-3²)=0`
`⇔(x+2)(x-2)(x+3)(x-3)=0`
`⇔`\(\left[ \begin{array}{l}x+2=0\\x-2=0\\x+3=0\\x-3=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-2\\x=2\\x=-3\\x=3\end{array} \right.\)
Vậy `S={-2;2;-3;3}`
`e)x^4-7x²+12=0`
`⇔x^4-3x²-4x²+12=0`
`⇔x²(x²-3)-4(x²-3)=0`
`⇔(x²-3)(x²-4)=0`
`⇔[x^2-(`$\sqrt[]{3}$ `)^2](x^2-2^2)=0`
`⇔(x+`$\sqrt[]{3}$`)(x-`$\sqrt[]{3}$`)(x+2)(x-2)=0`
`⇔`\(\left[ \begin{array}{l}x+\sqrt[]{3}=0\\x-\sqrt[]{3}=0\\x+2=0\\x-2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\sqrt[]{3}\\x=\sqrt[]{3}\\x=-2\\x=2\end{array} \right.\)
Vậy `S={-`$\sqrt[]{3}$`;`$\sqrt[]{3}$`;-2;2}`
`f)x^4+x²-6=0`
`⇔x^4+3x²-2x²-6=0`
`⇔x²(x²+3)-2(x²+3)=0`
`⇔(x²+3)(x²-2)=0`
Ta có:`x²≥0` với `∀x`
`⇒x²+3≥3>0`
`⇒` vô nghiệm
`⇔x²-2=0`
`⇔x²-(`$\sqrt[]{2}$ `)^2=0`
`⇔(x+`$\sqrt[]{2}$`)(x-`$\sqrt[]{2}$`)=0`
`⇔`\(\left[ \begin{array}{l}x+\sqrt[]{2}=0\\x-\sqrt[]{2}=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=-\sqrt[]{2}\\x=\sqrt[]{2}\end{array} \right.\)
Vậy `S={-`$\sqrt[]{2}$`;`$\sqrt[]{2}$`}`
