[tex]\displaystyle \sf 8\cdot (2x^2-5x+1)^2+9\cdot (2x^2-5x+2 ) -8=0 \\\\ 8\cdot (2x^2-5x+1)^2+9\cdot (2x^2-5x+1 + 1 ) -8=0 \\\\ \text{Fa{\c c}amos }(2x^2-5x+1) = y \ ,\ da{\'i}} :\\\\ 8\cdot y^2+9\cdot (y+1)-8 = 0 \\\\ 8\cdot y^2+9\cdot y+9-8=0 \\\\ 8\cdot y^2+9\cdot y+1=0 \\\\ y = \frac{-9\pm\sqrt{9^2-4\cdot 8\cdot 1}}{2\cdot 8 } \\\\\\ y = \frac{-9\pm\sqrt{49}}{16} \to y = \frac{-9\pm 7}{16} \\\\\\ y = \frac{-9-7}{16} \to \boxed{ \sf y = -1 }\\\\\\[/tex]
[tex]\displaystyle \sf y = \frac{-9+7}{16} \to y= \frac{-2}{16} \to \boxed{\sf y=\frac{-1}{8}}[/tex]
Desfazendo a troca de variável :
[tex]\displaystyle \sf \underline{\sf y = -1} \ :\ 2x^2-5x+1 = -1 \\\\ 2x^2-5x+2 = 0 \\\\ x =\frac{-(-5)\pm\sqrt{(-5)^2-4\cdot 2\cdot 2}}{2\cdot 2} \\\\\\ x= \frac{5\pm\sqrt{25-16}}{4} \to x= \frac{5\pm\sqrt{9}}{4} \to x = \frac{5\pm 3}{4} \\\\\\ \left\{ \begin{array}{I}\displaystyle \sf x = \frac{5+3}{4} \to \boxed{\sf x = 2 } \\\\ \displaystyle \sf x=\frac{5-3}{4} \to \boxed{\sf x = \frac{1}{2}} \end{array } \right[/tex]
[tex]\displaystyle \sf \underline{\sf y = \frac{-1}{8}} : \ 2x^2-5x+1=\frac{-1}{8} \\\\\\ 2x^2-5x+1+\frac{1}{8}= 0 \\\\\ 2x^2 -5x+\frac{9}{8} = 0 \\\\\\ x = \frac{-(-5)\pm\sqrt{\displaystyle (-5)^2-4\cdot 2\cdot \frac{9}{8}}}{\displaystyle 2\cdot 2} \\\\\\ x =\frac{5\pm\sqrt{25-9}}{4} \to x = \frac{5\pm\sqrt{16}}{4} \to x=\frac{5\pm4}{4} \\\\\\[/tex]
[tex]\left\{ \begin{array}{I} \displaystyle \sf x = \frac{5+4}{4}\to \boxed{\sf x=\frac{9}{4}} \\\\ \displaystyle \sf x=\frac{5-4}{4}\to \boxed{\sf x= \frac{1}{4}} \end{array} \right[/tex]
Portanto o Conjunto solução é :
[tex]\boxed{\displaystyle \sf S = \left\{x =2, \ x=\frac{1}{2},\ x= \frac{1}{4},\ x= \frac{9}{4}\ \right\}}\checkmark[/tex]
