Zadanie 4.
[tex]a) \\(x+2)^2-(x-2)^2=56\\x^2+4x+4-(x^2-4x+4)=56\\x^2+4x+4-x^2+4x-4=56\\8x=56 /:8\\x=7\\\\b) \\(x-3)^2-(4-x)^2=13\\x^2-6x+9-(16-8x+x^2)=13\\x^2-6x+9-16+8x-x^2=13\\2x-7=13 /+7\\2x=20 /:2\\x=10\\[/tex]
Zadanie 5.
[tex]a) \\9(x+1)^2-(3x-1)^2\geq -4\\9(x^2+2x+1)-(9x^2-6x+1)\geq-4\\9x^2+18x+9-9x^2+6x-1\geq-4\\24x+8\geq-4 /-8\\24x\geq-12 /:24\\x\geq-\frac{12}{24}\\x\geq-\frac12[/tex]
x∈<-¹/₂; ∞)
[tex]b)\\4(\frac12x-2)^2<(1-x)^2+7\\4(\frac14x^2-2x+4)<1-2x+x^2+7\\x^2-8x+16\frac86\\x>\frac43[/tex]
x∈(⁴/₃; ∞)
Zadanie 6
[tex]a)\\p=(3-\sqrt2)^2+(2\sqrt2-3)^2\\p=9-2*3*\sqrt2+2+(2\sqrt2)^2-2*2\sqrt2*3+9\\p=9-6\sqrt2+2+8-12\sqrt2+9\\p=28-18\sqrt2\\\text{Odp. p nie jest liczba wymierna}\\\\b)\\p=(\sqrt5-\sqrt3)^2+(\sqrt5+\sqrt3)^2\\p=5-2\sqrt{15}+3+5+2\sqrt{15}+3\\p=16\\\text{Odp. p jest liczba wymierna}[/tex]
Zadanie 7
[tex]a) \\q=\frac1{\sqrt2+1}-\frac{1}{\sqrt2-1}=\frac{1(\sqrt2-1)}{(\sqrt2+1)(\sqrt2-1)}-\frac{1(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)}=\frac{\sqrt2-1-(\sqrt2+1)}{2-1}=\frac{\sqrt2-1-\sqrt2-1}1=\frac{-2}1=-2\\\\b) \\q=\frac1{2-\sqrt3}+\frac6{\sqrt3+3}=\frac{\sqrt3+3+6(2-\sqrt3)}{(2-\sqrt3)(\sqrt3+3)}=\frac{\sqrt3+3+12-6\sqrt3}{2\sqrt3+6-3-3\sqrt3}=\frac{15-5\sqrt3}{3-\sqrt3}=\frac{15-5\sqrt3}{3-\sqrt3}*\frac{3+\sqrt3}{3+\sqrt3}=\frac{(15-5\sqrt3)(3+\sqrt3)}{9-3}=\frac{45+15\sqrt3-15\sqrt3-15}{6}=\frac{30}6=5[/tex]
