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125 (2,4)

$\sqrt{x+2} =2+ \sqrt{x-6} &amp;#10;\\\&amp;#10; \left \{ {{x+2 \geq 0} \atop {x-6 \geq 0}} \right. \Rightarrow x \geq 6&amp;#10;\\\&amp;#10;( \sqrt{x+2} )^2=(2+ \sqrt{x-6} )^2&amp;#10;\\\&amp;#10;x+2=4+4 \sqrt{x-6} +x-6&amp;#10;\\\&amp;#10;4 \sqrt{x-6} =4&amp;#10;\\\&amp;#10; \sqrt{x-6} =1&amp;#10;\\\&amp;#10;( \sqrt{x-6} )^2=1^2&amp;#10;\\\&amp;#10;x-6=1&amp;#10;\\\&amp;#10;x=7$
Ответ: 7

$\sqrt{22-x} - \sqrt{10-x} =2&amp;#10;\\\&amp;#10; \left \{ {{22-x \geq 0} \atop {10-x \geq 0}} \right. \Rightarrow x \leq 10&amp;#10;\\\&amp;#10; \sqrt{22-x} =2+ \sqrt{10-x} &amp;#10;\\\&amp;#10;( \sqrt{22-x} )^2=(2+ \sqrt{10-x})^2&amp;#10;\\\&amp;#10;22-x=4+4 \sqrt{10-x} +10-x&amp;#10;\\\&amp;#10;4 \sqrt{10-x} =8&amp;#10;\\\&amp;#10; \sqrt{10-x} =2&amp;#10;\\\&amp;#10;( \sqrt{10-x} )^2=2^2&amp;#10;\\\&amp;#10;10-x=4&amp;#10;\\\&amp;#10;x=6$
Ответ: 6