Xét hàm số `y=f(x)=(3-2\sqrt{2})x+\sqrt{2}`
`a,`
`f(0)=(3-2\sqrt{2}).0+\sqrt{2}=0+\sqrt{2}=\sqrt{2}`
`f(1)=(3-2\sqrt{2}).1+\sqrt{2}=3-2\sqrt{2}+\sqrt{2}=3-\sqrt{2}`
`f(3+2\sqrt{2})=(3-2\sqrt{2}).(3+2\sqrt{2})+\sqrt{2}=3^2-(2\sqrt{2})^2+\sqrt{2}=9-8+\sqrt{2}=1+\sqrt{2}`
Vậy `f(0)=\sqrt{2},f(1)=3-\sqrt{2},f(3+2\sqrt{2})=1+\sqrt{2}`
`b,`
`f(x)=0`
`⇔(3-2\sqrt{2})x+\sqrt{2}=0`
`⇔(3-2\sqrt{2})x=-\sqrt{2}`
`⇔x=-4-3\sqrt{2}`
Vậy `f(x)=0` khi `x=-4-3\sqrt{2}`
