Question

Найдите корень уравнения (3х-2)*(х+4)-3*(х+5)*(х-1)=0

Answer 1

$(3x-2)(x+4)-3(x+5)(x-1)=0$

 

$3x^{2}+12x-2x-8-3(x^{2}-x+5x-5)=0$

 

$3x^{2}+12x-2x-8-3x^{2}+3x-15x+15=0$

 

$(3x^{2}-3x^{2})+(12x-2x+3x-15x)+(-8+15)=0$

 

$-2x+7=0$

 

$2x=7$

 

$x=7:2$

 

$x=3,5$

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$(3cdot3,5-2)cdot(3,5+4)-3cdot(3,5+5)cdot(3,5-1)=0$

 

$(10,5-2)cdot7,5-3cdot8,5cdot2,5=0$

 

$8,5cdot7,5-63,75=0$

 

$63,75-63,75=0$

 

$0=0$