<em>√0.25+</em> <em>√</em> 36=0.5+6=6.5
<em>√</em> 8-2*(-4)=<em>√8*8=<em>√64=8</em></em>
<span><span><span><em><em><em>√</em> 8-2*2=2</em></em></span></span></span>
<span><span><span><em><em><em>√</em> 8-2*3.5=<em><em><em>√1=1</em></em></em></em></em></span></span></span>
<span><span><span><em><em>Ответ:6.5; <em><em>8; 2; 1</em></em></em></em></span></span></span>
В данной точке производные не существуют т.к. и в производной по х и в производной по у будет деление на (х+у), что=0
\left \{ {{\frac{2x-5}{4}}-2 \leq \frac{3-x}{4} \atop { \frac{5x+1}{5} \geq \frac{4-x}{4} }} \right.
\left \{ {{ \frac{2x-5-8-3+x}{4} \leq 0 } \atop { \frac{20x+5x-16}{20} \geq 0 }} \right.
\left \{ {{ \frac{3x-16}{20} \leq 0 } \atop { \frac{25x-16}{20} }} \right. \geq 0
\left \{ {{x= \frac{16}{25} } \atop {x= \frac{16}{3} }} \right.
\left \{ {{ \frac{10x-1}{3}}- \frac{2-5x}{4} \leq \frac{5-3x}{6} \atop { \frac{10x-1}{3}- \frac{3+7x}{4}- \frac{5-4x}{5} }} \right.
\left \{ {{ \frac{20x+10-15-35x+20+20x}{20} \geq 0 } \atop { \frac{40x-4-6-15x-10+6x}{12} \leq 0 }} \right.
\left \{ {{ \frac{5x+15}{20} \geq 0 } \atop { \frac{31x-20}{12} \leq 0 }} \right.
\left \{ { x= -\frac{20}{31}} \atop x= {3}} \right.
...