Найдите значение выражения (x−4)2+(x−12)2\sqrt{(x-4)^2}+\sqrt{(x-12)^2}(x−4)2+(x−12)2 при 4≤x≤124\leq x\leq124≤x≤12▸Ответ8▸Решение(x−4)2+(x−12)2=\sqrt{(x-4)^2}+\sqrt{(x-12)^2}=(x−4)2+(x−12)2=∣x−4∣+∣x−12∣\left | x-4 \right |+\left |x-12 \right |∣x−4∣+∣x−12∣ Т.к. 4≤x≤12⇒4\leq x\leq12\Rightarrow 4≤x≤12⇒ ∣x−4∣=x−4;∣x−12∣=12−x\left | x-4 \right |=x-4; \left|x-12 \right |=12-x∣x−4∣=x−4;∣x−12∣=12−x ∣x−4∣+∣x−12∣=\left | x-4 \right |+\left |x-12 \right |=∣x−4∣+∣x−12∣=x−4+12−x=12−4=8x-4+12-x=12-4=8x−4+12−x=12−4=8