Question

1. sqrt10{(x - 2)^{10}}, если x geqslant 2.

Explanation:

Step1: Recall the nth root property

For an even root \(\sqrt[n]{a^n}\), when \(n\) is even, \(\sqrt[n]{a^n}=\vert a\vert\). Here, \(n = 10\) (even) and \(a=(x - 2)\).

Step2: Apply the condition \(x\geq2\)

Since \(x\geq2\), then \(x - 2\geq0\). The absolute value of a non - negative number is the number itself, so \(\vert x - 2\vert=x - 2\) when \(x\geq2\). Thus, \(\sqrt[10]{(x - 2)^{10}}=\vert x - 2\vert=x - 2\) (because \(x\geq2\) implies \(x - 2\geq0\)).

Answer:

\(x - 2\)