tìm y sao cho y >12 : 12.[(y.0,5:10-y)+1]=[(y.0,5:10.y).0,5].y

Ta có

$12[ (\dfrac{1}{20} y - y) + 1] = [(\dfrac{1}{20} . y . y).0,5]y$

$<-> 12 (-\dfrac{19}{20} y + 1) = \dfrac{1}{40} y^2 . y$

$<->-\dfrac{57}{5} y + 12 = \dfrac{1}{40} y^3$

$<-> -456 y + 480 = y^3$

$<-> y^3 + 456y - 480 = 0$

Ptrinh này ko có nghiệm $y>12$ nên ko có số $y$ nào thỏa mãn.

12[(120y−y)+1]=[(120.y.y).0,5]y12[(120y−y)+1]=[(120.y.y).0,5]y

<−>12(−1920y+1)=140y2.y<−>12(−1920y+1)=140y2.y

<−>−575y+12=140y3<−>−575y+12=140y3

<−>−456y+480=y3<−>−456y+480=y3

<−>y3+456y−480=0