Question

Помогите решить (6и7)

Answer 1

$6)\; \; \left \{ {{x^2+4y=-53} \atop {7x+2y=-2}} \right. \; \left \{ {{x^2+4(-3,5x-1)=-53} \atop {y=-3,5x-1}} \right. \; \left \{ {{x^2-14x+49=0} \atop {y=-3,5x-1}} \right. \\\\x^2-14x+49=(x-7)^2=0\; \; \Rightarrow \; \; x=7\\\\y=-3,5\cdot 7-1\; ,\; \; y=-25,5\\\\Otvet:\; \; (7\, ;\, -25,5)\, .$

$7)\; \; \Big (\frac{3}{\sqrt5+\sqrt8}+\frac{3}{\sqrt8+\sqrt{11}}-\sqrt5\Big )^2=\Big (\frac{3(\sqrt5-\sqrt8)}{5-8}+\frac{3(\sqrt8-\sqrt{11})}{8-11}-\sqrt5\Big )^2=\\\\=\Big (-(\sqrt5-\sqrt8)-(\sqrt8-\sqrt{11})-\sqrt5\Big )^2=\Big (-\sqrt5+\sqrt8-\sqrt8+\sqrt{11}-\sqrt5\Big )^2=\\\\=\Big (\sqrt{11}-2\sqrt5\Big )^2=11-4\sqrt5\cdot \sqrt{11}+4\cdot 5=31-4\sqrt{55}\\\\\\P.S.\; \; (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$