Đáp án:
Giải thích các bước giải:
a)$\sqrt[]{x-6}$ $=13$
⇔$\sqrt[]{(x-6)^{2}}$ $=13^{2}$
⇔$x-6=169$
⇔$x=175$
b)$\sqrt[]{x^{2}-2x+4}$ $=x-1$
⇔$x^{2}$ $-2x+4=$ $x^{2}$ $-2x+1$
⇔$4=1$ (vô lí)
c)$\sqrt[]{x^{2}-8x+16}$ $=9x-1$
⇔$\sqrt[]{(x^{2}-8x+16)^2}$ $=(9x-1)^2$
⇔$x^{2}-8x+16$ $=81x^{2}$ $-18x+1$
⇔$-80x^{2}$ $+10x+15$=0
⇔\(\left[ \begin{array}{l}x=\frac{1}{2}\\x=-\frac{3}{8}\end{array} \right.\)
d)$\sqrt[]{x^{2}-x-4}$ $=$ $\sqrt[]{x-1}$
⇔$\sqrt[]{(x^{2}-x-4)^2}$ $=$ $\sqrt[]{(x-1)^2}$
⇔$x^{2}$ $-x-4$ $=$ $x-1$
⇔$x^{2}$ $-2x-3$ $= 0$
⇔\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
e)$\sqrt[]{x^2-4x+4}$ $=$ $\sqrt[]{4x^2-12x+9}$
⇔$\sqrt[]{(x-2)^2}$ $=$ $\sqrt[]{(2x-3)^2}$
⇔$x-2 =$ $2x-3$
⇔$x=1$
g)$\sqrt[]{x+2\sqrt[]{x-1}}$ $= 2$
⇔$\sqrt[]{(x+\sqrt[]{4x-4})^2}$ $= 2^2$
⇔$\sqrt[]{4x-4}$ $= 4-x$
⇔ \(\left[ \begin{array}{l}x=2\\x=10\end{array} \right.\)
