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rozwiaz rownanie (3-x)²-(x+1/3)²=2/9 dam naj​

Odpowiedź :

LoczeeQviz

[tex](3 - x) ^{2} - (x + \frac{1}{3} )^{2} = \frac{2}{9} \\ (3 - x)(3 - x) - (x + \frac{1}{3} )(x + \frac{1}{3} ) = \frac{2}{9} \\ 9 - 3x - 3x + {x}^{2} - {x}^{2} + \frac{1}{3} x + \frac{1}{3} x + \frac{1}{9} = \frac{2}{9} \\ - 6x + \frac{2}{3} x = \frac{2}{9} - \frac{1}{9} - 9 \\ - 5 \frac{1}{3} x = \frac{1}{9} - 9 \\ - \frac{16}{3} x = - 8 \frac{8}{9} \\ - \frac{16}{3} x = - \frac{80}{9} \\ x = ( - \frac{80}{9} ) \div ( - \frac{16}{3} ) \\ x = ( - \frac{80}{9} ) \times ( - \frac{3}{16} ) \\ x = (- \frac{80}{3} ) \times ( - \frac{1}{16} ) \\ x = \frac{80}{48} \\ x = 1 \frac{32}{48} \\ x = 1 \frac{2}{3} [/tex]

Myślę że pomogłem ;)

Krysiaviz

[tex](3-x)^2-(x+ \frac{1}{3})^2= \frac{2}{9}\\\\9-6x+x^2-(x^2+\frac{2}{3}x+\frac{1}{9})=\frac{2}{9}\\\\9-6x+x^2- x^2-+\frac{2}{3}x-\frac{1}{9} =\frac{2}{9}\\\\ -6x+x^2- x^2- \frac{2}{3}x =\frac{2}{9}+\frac{1}{9}-9\\\\-6\frac{2}{3}x=\frac{3}{9}-9\\\\- \frac{20}{3}x=\frac{1}{3}-9\\\\ - \frac{20}{3}x=-8\frac{2}{3}[/tex]

[tex]- \frac{20}{3}x=- \frac{26}{3} \ \ |*(-\frac{3}{20})\\\\x=\frac{\not{26}^{13}}{\not{3}^1}*\frac{\not{3}^1}{\not{20}^{10}}\\\\x=\frac{13}{10}=1\frac{3}{10}[/tex]