4=(23/4-3)+(-15/(4x-3)^2)

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Solution for 4=(23/4-3)+(-15/(4x-3)^2) equation:


D( x )

(4*x-3)^2 = 0

(4*x-3)^2 = 0

(4*x-3)^2 = 0

4*x-3 = 0 // + 3

4*x = 3 // : 4

x = 3/4

x in (-oo:3/4) U (3/4:+oo)

4 = 23/4-15/((4*x-3)^2)-3 // - 23/4-15/((4*x-3)^2)-3

3-(-15/((4*x-3)^2))-(23/4)+4 = 0

15*(4*x-3)^-2-23/4+3+4 = 0

15/((4*x-3)^2)-23/4+3+4 = 0

(4*15)/(4*(4*x-3)^2)+(-23*(4*x-3)^2)/(4*(4*x-3)^2)+(3*4*(4*x-3)^2)/(4*(4*x-3)^2)+(4^2*(4*x-3)^2)/(4*(4*x-3)^2) = 0

3*4*(4*x-3)^2-23*(4*x-3)^2+4^2*(4*x-3)^2+4*15 = 0

192*x^2-368*x^2+256*x^2+552*x-288*x-384*x-147+108+144 = 0

256*x^2-176*x^2+264*x-384*x-39+144 = 0

80*x^2-120*x+105 = 0

80*x^2-120*x+105 = 0

5*(16*x^2-24*x+21) = 0

16*x^2-24*x+21 = 0

DELTA = (-24)^2-(4*16*21)

DELTA = -768

DELTA < 0

5 = 0

5/(4*(4*x-3)^2) = 0

5/(4*(4*x-3)^2) = 0 // * 4*(4*x-3)^2

5 = 0

x belongs to the empty set

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