<s>[INST] let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$ let $a-b=6$ and $b\leq-1$ and $a\leq-2$. Find $A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}$
My Try $$A=\sqrt{a^2+4a+4}+\sqrt{b^2+2b+1}=$$then I can't
Please just a hint.
 [/INST] Simply write $b$ in terms of $a$ (and mind the conditions on $a$ and $b$:  $a \leq −2$ so $a+2\leq0$ and $b\leq-1$ so $b+1\leq0$) and you get
$$A=\sqrt{(a+2)^2}+\sqrt{(b+1)^2}=\sqrt{(a+2)^2}+\sqrt{(a-6+1)^2}=|a+2| + |a-5|=-a-2-a+5=-2a+3$$
since $a+2\leq0$ and $b+1\leq0$ ($\Rightarrow |a+2| = -a-2 \text{ and } |a-5| = -a-5$).
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