Đáp án:
Giải thích các bước giải:
$\sqrt[]{9x^2-12x+4-2016}=1$
⇔$9x^2-12x+4-2016=1_{}$
⇔ $(3x-2)^2-2017=0_{}$
⇔$(3x-2-_{}$ $\sqrt[]{2017})(3x-2+$ $\sqrt[]{2017})=0$
⇔\(\left[ \begin{array}{l}3x-2-\sqrt[]{2017}=0\\3x-2+\sqrt[]{2017}=0\end{array} \right.\)
$TH1:3x-2-\sqrt[]{2017}=0_{}$
⇔$3x=_{}$ $\sqrt[]{2017}+2$
⇔$x=\frac{\sqrt[]{2017}+2}{3}$
$TH2: 3x-2+\sqrt[]{2017}=0_{}$
⇔$3x=_{}$ $-\sqrt[]{2017}+2$
⇔$x=\frac{-\sqrt[]{2017}+2}{3}$
