Odpowiedź:
Rozwiazania z rozpisaniem:
[tex]1. \\(3+\sqrt2)^2=3^2+2*3*\sqrt2+(\sqrt2)^2=9+6\sqrt2+2=11+6\sqrt2\\\\2. \\(6-\sqrt{11})^2=6^2-2*6*\sqrt{11}+(\sqrt{11})^2=36-12\sqrt{11}+11=47-12\sqrt{11}\\\\3. (\sqrt{101}+10)(\sqrt{101}-10)=(\sqrt{101})^2-10^2=101-100=1\\\\4. (5\sqrt2+7)(7-5\sqrt2)=(7-5\sqrt2)(5\sqrt2+7)=-(-7+5\sqrt2)(5\sqrt2+7)=-(5\sqrt2-7)(5\sqrt2+7)=-((5\sqrt2)^2-7^2)=-(50-49)=-1[/tex]
[tex]5. \\(\sqrt{51}+2\sqrt3)(\sqrt{51}-2\sqrt3)=(\sqrt{51})^2-(2\sqrt3)^2=51-12=39\\\\6. \\(\sqrt{2\sqrt2-\sqrt5}-\sqrt{\sqrt5+2\sqrt2})^2=\\(\sqrt{2\sqrt2-\sqrt5}-\sqrt{2\sqrt2+\sqrt5})^2=\\(\sqrt{2\sqrt2-\sqrt5})^2-2(\sqrt{2\sqrt2-\sqrt5})(\sqrt{2\sqrt2+\sqrt5})+(\sqrt{2\sqrt2+\sqrt5})^2=\\2\sqrt2-\sqrt5-2\sqrt{(2\sqrt2-\sqrt5)(2\sqrt2+\sqrt5)}+2\sqrt2+\sqrt5=\\4\sqrt2-2\sqrt{(2\sqrt2)^2-(\sqrt5)^2}=\\4\sqrt2-2\sqrt{8-5}=\\4\sqrt2-2\sqrt3[/tex]
[tex]7. \\(\sqrt{\sqrt6+\sqrt3}+\sqrt{\sqrt6-\sqrt3})^2=\\(\sqrt{\sqrt6+\sqrt3})^2+2(\sqrt{\sqrt6+\sqrt3})(\sqrt{\sqrt6-\sqrt3})+(\sqrt{\sqrt6-\sqrt3})^2=\\\sqrt6+\sqrt3+2\sqrt{(\sqrt6+\sqrt3)(\sqrt6-\sqrt3)}+\sqrt6-\sqrt3=\\2\sqrt6+2\sqrt{(\sqrt6)^2-(\sqrt3)^2}=\\2\sqrt6+2\sqrt{6-3}=\\2\sqrt6+2\sqrt3[/tex]
[tex]8. \\(4y+5)^2=(4y)^2+2*4y*5+5^2=16y^2+40y+25\\\\9. \\(3-6a)^2=3^2-2*3*6a+(6a)^2=9-36a+36a^2\\\\10. \\(-3-s)^2=(-3)^2-2*(-3)*s+s^2=9+6s+s^2\\\\11. \\(-3-2y)^2=(-3)^2-2*(-3)*2y+(2y)^2=9+12y+4y^2[/tex]
Szczegółowe wyjaśnienie:
Wzory ponizej:
[tex](a+b)^2=a^2+2ab+b^2\\(a-b)^2=a^2-2ab+b^2\\(a+b)(a-b)=a^2-b^2[/tex]
