Question

за решение +баллы и сердечко поставлю

Answer 1

Смотри...................................

Answer 2

Ответ:

Объяснение:

х$x^{2} *(x-5)-(x-5)=0\\(x-5)*(x^{2} -1)=0\\x-5=0\\x1=5\\x^{2} -1=0\\x^{2} =1\\x2=1\\x3=(-1)$

Б)Пусть х^2 = t, то х^4=t^2

$t^{2} -2t-8=0$

D=4-4*1*(-8)=4+32=36

t1=$\frac{2-6}{2} =-2$

t2=$\frac{2+6}{2}=4$

$x^{2} =-2\\x1=$нет решения

x$x^{2} =4\\x1=2\\x2=-2$

B)

$x^{4} -5x^{3} +10x^{2} -5x^{3} +25x^{2} -50x+24=0\\x^{4} -10x^{3} +35x^{2} -50x+24=0\\x^{4} -x^{3} -9x^{3} +9x^{2} +26x^{2} -26x-24x+24=0\\x^{3}*( x-1) -9x^{2}*( x-1) +26x*( x-1) -24*( x-1) =0\\( x-1)*(x^{3}-9x^{2} +26x -24) =0\\( x-1)*(x^{2} *(x-2)-7x*(x-2)+12 *(x-2))= 0\\( x-1)*(x-2) *(x*(x-3)-4*(x-3))= 0\\( x-1)*(x-2) *(x-3)*(x-4)= 0\\x1=1\\x2=2\\x3=3\\x4=4$