Question

допоможіть будь ласка ​

Answer 1

Пояснення:

$(x-1)(x-7)((x-4)(x+2)=40\\\\(x-1)(x-4)*(x-7)*(x+2)=40\\\\(x^2-x-4x+4)*(x^2+2x-7x-14)=40\\\\(x^2-5x+4)*(x^2-5x-14)=40$

Нехай x²-5x=t     ⇒

$\displaystyle\\(t+4)*(t-14)=40\\\\t^2+4t-14t-56=40\\\\t^2-10t-96=0\\\\t^2-16t+6t-96=0\\\\t*(t-16)+6*(t-16)=0\\\\(t-16)*(t+6)=0\\\\t-16=0\\\\t_1=16\\\\t+6=0\\\\t_2=-6.\\\\1.\\t=16\ \ \ \ \Rightarrow\\\\x^2-5x=16\\\\x^2-5x-16=0\\\\D=5^2-4*1*(-16)=25+64=89\\\\\sqrt{D}=\sqrt{89}\\\\x_{1,2} =\frac{5б\sqrt{89} }{2}$

2.

t=-6      ⇒

$x^2-5x=-6\\\\x^2-5x+6=0\\\\x^2-3x-2x+6=0\\\\x*(x-3)-2*(x-3)=0\\\\(x-3)*(x-2)=0\\\\x-3=0\\\\x_3=3.\\\\x-2=0\\\\x_4=2.$

Відповідь: х₁,₂=(5±√89)/2, х₃=3, х₄=2.