Question

помогите, пожалуйста!!
можете просто первый вариант решить​

Answer 1

$\displaystyle\bf\\1)\\\\\sqrt{2007^{2} -2006\cdot 2008}=\sqrt{2007^{2} -(2007-1)\cdot(2007+1)} =\\\\=\sqrt{2007^{2} -(2007^{2} -1^{2}) } =\sqrt{2007^{2} -2007^{2} +1} =\sqrt{1} =1\\\\\\2)\\\\\sqrt{2008^{2} -2003\cdot 2013}=\sqrt{2008^{2} -(2008-5)\cdot(2008+5)} =\\\\=\sqrt{2008^{2} -(2008^{2} -5^{2}) } =\sqrt{2008^{2} -2008^{2} +25} =\sqrt{25} =5\\\\\\3)\\\\\sqrt{2009^{2} -2003\cdot 2015}=\sqrt{2009^{2} -(2009-6)\cdot(2009+6)} =$

$\displaystyle\bf\\=\sqrt{2009^{2} -(2009^{2} -6^{2}) } =\sqrt{2009^{2} -2009^{2} +36} =\sqrt{36} =6$