(3y3-12y2-y+4)/ 9y4-1 =0

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вопрос

To solve for y, first factor out the numerator and denominator:

(3y^3 - 12y^2 - y + 4) / (9y^4 - 1) = 0

Numerator: (3y^3 - 12y^2 - y + 4) = y^2(3y - 12) - 1(3y - 4) = (y^2 - 1)(3y - 4)

Denominator: (9y^4 - 1) = (3y^2 + 1)(3y^2 - 1) = (3y^2 + 1)(√3y - 1)(√3y + 1)

Now, the equation becomes:

((y^2 - 1)(3y - 4)) / ((3y^2 + 1)(√3y - 1)(√3y + 1)) = 0

Setting the numerator equal to 0:

(y^2 - 1)(3y - 4) = 0

This equation can be solved by setting each factor equal to 0:

y^2 - 1 = 0
y^2 = 1
y = ±1

3y - 4 = 0
3y = 4
y = 4/3

Hence, the possible solutions for the equation are y = 1, y = -1, y = 4/3.