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(√6+2)^2+(√3-2√2)^2-(√10-√3)(√10+√3)

Răspuns :

[tex]\displaystyle \left( \sqrt{6} +2 \right)^2+\left( \sqrt{3} -2 \sqrt{2} \right)^2-\left( \sqrt{10} - \sqrt{3} \right)\left( \sqrt{10} + \sqrt{3} \right)= \\ \\ =\left( \sqrt{6} \right)^2+2 \cdot \sqrt{6} \cdot 2+2^2+\left( \sqrt{3} \right)^2-2 \cdot \sqrt{3} \cdot 2 \sqrt{2} + \left(2 \sqrt{2} \right)^2-(10-3)= \\ \\ =6+4 \sqrt{6} +4+3-4 \sqrt{6} +8-10+3=14[/tex]