Per favore, mi potete svolgere questa equazione???

[Studenteimpazzito]

Ecco qua l'equazione scannerizzata http://farm9.staticflickr.com/8445/7942813488_a0e561e6ec_k.jpg
Grazie mille in anticipo.

[bimbozza]

[math](\frac{2}{3} )^2 \frac{x+ \frac{1}{5}}{2- \frac{4}{3}} \frac{3}{19}[ \frac{2}{5}: (1- \frac{3}{4})- \frac{1}{3}]- \frac{(x+ \frac{2}{3})^2}{ \frac{1}{6}}+ \frac{(x+ \frac{1}{3})(x-\frac{1}{3})}{1- \frac{5}{6}}+ \frac{2}{3}=0[/math]

[math](\frac{2}{3} )^2 \frac{x+ \frac{1}{5}}{ \frac{6-4}{3}} \frac{3}{19}[ \frac{2}{5}: (\frac{4-3}{4})- \frac{1}{3}]- 6(x+ \frac{2}{3})^2+ \frac{x^2 -\frac{1}{9}}{\frac{6-5}{6}}+ \frac{2}{3}=0[/math]

[math](\frac{2}{3} )^2 \frac{x+ \frac{1}{5}}{ \frac{2}{3}} \frac{3}{19}[ \frac{2}{5}: (\frac{1}{4})- \frac{1}{3}]- 6(x+ \frac{2}{3})^2+ \frac{x^2 -\frac{1}{9}}{\frac{1}{6}}+ \frac{2}{3}=0[/math]

[math](\frac{2}{3} )^2 \frac{3(x+ \frac{1}{5})}{ 2} \frac{3}{19}[ \frac{2}{5}* 4 - \frac{1}{3}]- 6(x^2+ \frac{4}{9}+ \frac{4x}{3})+ 6(x^2 - \frac{1}{9})+ \frac{2}{3}=0[/math]

[math](\frac{2}{3} )^2 \frac{3(x+ \frac{1}{5})}{ 2} \frac{3}{19}[ \frac{8}{5}- \frac{1}{3}]- 6x^2- \frac{8}{3}- 8x+ 6x^2 - \frac{2}{3}+ \frac{2}{3}=0[/math]

[math](\frac{2}{3} )^2 \frac{3(x+ \frac{1}{5})}{ 2} \frac{3}{19} \frac{24-5}{15}- \frac{8}{3}- 8x =0[/math]

[math](\frac{2}{3} )^2 (\frac{3x}{2}+ \frac{3}{10}) \frac{3}{19} \frac{19}{15} -\frac{8}{3}- 8x =0[/math]

[math]\frac{2x}{15} + \frac{2}{75}- \frac{8}{3} - 8x =0[/math]

[math]\frac{2x-120x}{15} = \frac{-2+200}{75}[/math]

[math]\frac{-118x}{15} = \frac{198}{75}[/math]

[math]x =-\frac{198}{75} \frac{15}{118}[/math]

[math]x =-\frac{99}{295} [/math]