Если sqrt(8-t) - sqrt(3-t)=2, то sqrt(8-t) +sqrt(3-t) равно ?

[latex] \sqrt{8-t}- \sqrt{3-t}=2\\ ( \sqrt{8-t}- \sqrt{3-t})^2=2^2\\(8-t)+(3-t)-2 \sqrt{(8-t)(3-t)}=4\\11-2t-2 \sqrt{24-11t+t^2}=4\\7-2t=2 \sqrt{24-11t+t^2}\\(7-2t)^2=(2 \sqrt{24-11t+t^2})^2\\49+4t^2-28t=4(24-11t+t^2)\\4t^2-28t+49=4t^2-44t+96\\44t-28t=96-49\\16t=47\\t=47/16\\\\ \sqrt{8-t}+ \sqrt{3-t}=?\\ \sqrt{8- \frac{47}{16} }+ \sqrt{3- \frac{47}{16} }= \sqrt{\frac{128-47}{16}}+ \sqrt{ \frac{48-47}{16} }= \sqrt{ \frac{81}{16} }+ \sqrt{ \frac{1}{16} }= [/latex] [latex]=\frac{9}{4}+ \frac{1}{4}=\frac{10}{4}=2,5 [/latex] Ответ: 2,5