Odpowiedź:
[tex](x-\frac{4}{3})(x+\frac{4}{3})-(x-\frac{2}{3})^2-2x=2\\\\x^2-\frac{16}{9}-(x^2-\frac{4}{3}x+\frac{4}{9})-2x=2\\\\\not x^2-\frac{16}{9}-\not x^2+\frac{4}{3}x-\frac{4}{9}-2x=2\\\\-\frac{20}{9}+\frac{4}{3}x-2x=2\\\\\frac{4}{3}x-\frac{6}{3}x=2+\frac{20}{9}\\\\-\frac{2}{3}x=\frac{18}{9}+\frac{20}{9}\\\\-\frac{2}{3}x=\frac{38}{9}\ \ /:(-\frac{2}{3})\\\\x=\frac{\not38^1^9}{\not9_{3}}\cdot(-\frac{\not3^1}{\not2_{1}})\\\\x=-\frac{19}{3}\\\\x=-6\frac{1}{3} [/tex]
Szczegółowe wyjaśnienie:
[tex]Zastosowano\ \ wzory\ \ skr\'oconego\ \ mno\.zenia\\\\(a-b)(a+b)=a^2-b^2\\\\(x-\frac{4}{3})(x+\frac{4}{3})=x^2-(\frac{4}{3})^2=x^2-\frac{16}{9}\\\\\\(a-b)^2=a^2-2ab+b^2\\\\(x-\frac{2}{3})^2=x^2-2x\cdot\frac{2}{3}+(\frac{2}{3})^2=x^2-\frac{4}{3}x+\frac{4}{9} [/tex]
