Đáp án:
Giải thích các bước giải:
$\sqrt[]{x^2-2x+1}=2$
$⇔\sqrt[]{(x-1)^2}=2$
$⇔\left[ \begin{array}{l}x-1=2\\x-1=-2\end{array} \right.$
$⇔\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.$
$\sqrt[]{x^2-1}=x$$ĐKXĐ:x\neq±1$
$⇔x^2-1=x^2(vn)$
$x-5\sqrt[]{x-2}=-2$$ĐKXĐ:x\neq2$
$⇔-5\sqrt[]{x-2}=-x-2$
$⇔\sqrt[]{x-2}=\frac{2+x}5$
$⇔x-2=\frac{4+4x+x^2}{25}$
$⇔25(x-2)=x^2+4x+4$
$⇔25x-50-x^2-4x-4=0$
$⇔-x^2+21-54=0$
$⇔x^2-21+46=0$
$Δ=(-21)^2-4.1.54=225>0$
$⇔\left[ \begin{array}{l}x=\frac{21+\sqrt[]{225}}2=18\\x=\frac{21-\sqrt[]{225}}2=3\end{array} \right.$
Xin hay nhất!!!
