Verified answer

Ответ:

[tex]x_1=3, ~~x_2=-1,5[/tex]

Объяснение:

[tex]\displaystyle\frac{x^2}{18} +\frac{x+1}{12} =\frac{x+2}{6} ;\\\underbrace{\frac{x^2}{18} } _{2}+\underbrace{\frac{x+1}{12} } _{3}=\underbrace{\frac{x+2}{6} } _{6};\\2x^2+3x+3=6x+12;\\2x^2-3x-9=0;\\x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} =\frac{3\pm\sqrt{9-4*2*(-9)} }{2*2} =\frac{3\pm9}{4} ;\\x_1=\frac{3+9}{4} =3;\\x_2=\frac{3-9}{4} =-1,5[/tex]