Đáp án:
`a)x^2-6xy+16y^2-2y+1=0`
`<=>x^2-6xy+9y^2+7y^2-2y+1=0`
`<=>(x-3y)^2+y^2-2y+1+6y^2=0`
`<=>(x-3y)^2+(y-1)^2+6y^2=0`
Vì \(\begin{cases}(x-3y)^2 \ge 0\\(y-1)^2 \ge 0\\6y^2 \ge 0\\\end{cases}\)
`=>VT>=0`.
Dấu "=" xảy ra khi \(\begin{cases}x-3y=0\\y-1=0\\y=0\\\end{cases}\)
`<=>` \(\begin{cases}x=3y\\y=1\\y=0\\\end{cases}(\text{vô lý})\)
Vậy pt vô nghiệm.
`b)(x-2)(x+2)(x^2-10)=0`
`<=>` \(\left[ \begin{array}{l}x-2=0\\x+2=0\\x^2-10=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\\x^2=10\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\\x=\sqrt{10}\\x=-\sqrt{10}\end{array} \right.\)
Vậy `S={2,-2,\sqrt{10},-\sqrt{10}}`.
