Question

ДАЮ 50 БАЛЛОВ
Решите уравнение:
1) 25х^3 – 10х^2 + х = 0; 2) х^3 – 4х^2 – 9x + 36 = 0.

Answer 1

1.

$25x^3-10x^2 + x = 0$

$x*(25x^2-10x+1 ) = 0<=>\left \ [{{x=0} \atop {25x^{2} -10x+1=0}} \right.=>\left \ [{{x_1=0} \atop {(5x-1)^{2} =0}} \right.=>\left \ [{{x_1=0} \atop {x_2=x_3=0,2}} \right.$

Ответ: {0;  0,2}

2.

$x^3 - 4x^2- 9x + 36 = 0.$

$(x^3 - 4x^2)- (9x - 36) = 0.$

$x^{2} (x - 4)- 9(x - 4) = 0.$

$(x^{2} -9)(x - 4)=0$

$(x^{2} -3^2)(x - 4)=0$

$(x-3)(x+3)(x - 4)=0$

$x+3=0$    =>    $x_1=-3$

$x-3=0$     =>    $x_2=3$

$x-4=0$     =>    $x_3=4$

Ответ:   {-3;  3;  4}

Answer 2

Ответ:

$1)\ \ 25x^3-10x^2+x=0\\\\x\, (25x^2-10x+1)=0\ \ ,\ \ \ x\, (5x-1)^2=0\\\\x_1=0\ \ ,\\\\5x-1=0\ \ ,\ \ x_2=\dfrac{1}{5}=0,2\\\\Otvet:\ \ x_1=0\ ,\ x_2=0,2\ .$

$2)\ \ x^3-4x^2-9x+36=0\\\\x^2(x-4)-9(x-4)=0\\\\(x-4)(x^2-9)=0\\\\(x-4)(x-3)(x+3)=0\\\\Otvet:\ \ x_1=4\ ,\ x_2=3\ ,\ x_3=-3\ .$