Question

ktoś mi pomoże pls. ​

Answer 1

[tex]8.\\\\x=2-\sqrt{7}\ \ \ \ y=3+\sqrt{7}\\\\a)\ \ x\cdot y\\\\(2-\sqrt{7})(3+\sqrt{7})=2\cdot3+2\sqrt{7}-3\sqrt{7}-\sqrt{7}\cdot\sqrt{7}=6+2\sqrt{7}-3\sqrt{7}-7=\\\\=-1-\sqrt{7}\\\\\\b)\ \ \frac{x}{y}\\\\\\\dfrac{2-\sqrt{7}}{3+\sqrt{7}}=\dfrac{2-\sqrt{7}}{3+\sqrt{7}}\cdot\dfrac{3-\sqrt{7}}{3-\sqrt{7}}=\dfrac{(2-\sqrt{7})(3-\sqrt{7})}{(3+\sqrt{7})(3-\sqrt{7})}=\dfrac{2\cdot3-2\sqrt{7}-3\sqrt{7}-\sqrt{7}\cdot(-\sqrt{7})}{3^2-(\sqrt{7})^2}=[/tex]

[tex]=\dfrac{6-5\sqrt{7}+\sqrt{7}\cdot\sqrt{7}}{9-7}=\dfrac{6-5\sqrt{7}+7}{2}=\dfrac{13-5\sqrt{7}}{2}\\\\\\c)\ \ 3y-x^2\\\\3(3+\sqrt{7})-(2-\sqrt{7})^2=9+3\sqrt{7}-(2^2-2\cdot2\cdot\sqrt{7}+(\sqrt{7})^2)=\\\\=9+3\sqrt{7}-(4-4\sqrt{7}+7)=9+3\sqrt{7}-(11-4\sqrt{7})=9+3\sqrt{7}-11+4\sqrt{7}=\\\\=-2+7\sqrt{7}[/tex]

[tex]d)\ \ x^2-y^2\\\\(2-\sqrt{7})^2-(3+\sqrt{7})^2=2^2-2\cdot2\cdot\sqrt{7}+(\sqrt{7})^2-(3^2+2\cdot3\cdot\sqrt{7}+(\sqrt{7})^2)=\\\\=4-4\sqrt{7}+7-(9+6\sqrt{7}+7)=11-4\sqrt{7}-(16+6\sqrt{7})=\\\\=11-4\sqrt{7}-16-6\sqrt{7}=-5-10\sqrt{7}[/tex]

[tex]9.\\\\a)\\\\(x+3)(x-3)-(x+4)^2+(5x-1)^2=\\\\=x^2-3^2-(x^2+2x\cdot4+4^2)+(5x)^2-2\cdot5x\cdot1+1^2=\\\\x^2-9-(x^2+8x+16)+25x^2-10x+1=\\\\=x^2-9-x^2-8x-16+25x^2-10x+1=25x^2-18x-24\\\\\\b)\\\\-(-3x-5)^2-(-x-6)^2=\\\\=-((-3x)^2-2\cdot(-3x)\cdot5+5^2)-((-x)^2-2\cdot(-x)\cdot6+6^2)=\\\\=-(9x^2+30x+25)-(x^2+12x+36)=-9x^2-30x-25-x^2-12x-36=\\\\=-10x^2-42x-61[/tex]

[tex]Zastosowano\ \ wzory\ \ skr\'oconego\ \ mno\.zenia\\\\(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]