<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] Hint: You need to know that $a\leqslant -2$ iff $a+2\leqslant 0$, and $b\leqslant -1$ iff $b+1\leqslant 0$. And the most important thing: $$ \color{lightblue}{\boxed{\color{white}{\boxed{\color{white}{\overline{\underline{\color{black}{\displaystyle\sqrt{\rm X^2}=|\,\rm X|\,}}}}}}}}$$
I hope this helps. 
Best wishes, $\mathcal H$akim.
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