Question

(х-3)³-(х-5)²(х+1)+4=0

Answer 1

Х-3(3)-х-5(2)*х+1+4
х(3)*9-х(2)*10*х+1+4=0
х(3)*х(2)*х=9+10+1+4
х(7)=24

Answer 2

X³-27-9x(x-3)-(x²-10x+25)(x+1)+4=0
x³-27-9x²+27x-(x³+x²-10x²-10x+25x+25)+4=0
x³-27-9x²+27x-x³-x²+10x²+10x-25x-25+4=0
12x-48=0
12x=48
x=4

Answer 3

Let's simplify the expression step by step:

(х-3)³ - (х-5)²(х+1) + 4 = 0

Expand the first term using the formula for the cube of a binomial:

(х³ - 9х² + 27х - 27) - (х² - 6х + 25)(х + 1) + 4 = 0

Distribute the second term:

(х³ - 9х² + 27х - 27) - (х³ + 19х² - 19х - 25) + 4 = 0

Simplify by combining like terms:

-10х² + 43х - 48 = 0

Now we can solve for x using the quadratic formula:

х = (-b ± sqrt(b² - 4ac)) / 2a

where a = -10, b = 43, and c = -48

х = (-43 ± sqrt(43² - 4(-10)(-48))) / 2(-10)

х = (-43 ± sqrt(2333)) / (-20)

So we have two solutions:

х ≈ 0.435

х ≈ 4.388