4)x+$\sqrt{5+\sqrt{x-1}}$=6 (Đk:x$\geq$1)
$\Leftrightarrow$x-1+$\sqrt{5+\sqrt{x-1}}$=5 (1)
Đặt t=$\sqrt{x-1}$$\geq$0
(1)$\Leftrightarrow$$t^{2}+\sqrt{5+t}=5$
Đặt y=$\sqrt{5+t}$$\geq$0
Ta có hpt
$\left\{\begin{matrix} y^{2}=5+t\\t^{2}=5-y \end{matrix}\right.$
$\Rightarrow$$(y-t)(y+t)=y+t$
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