Question

решить с помощью замены (x-1)^2-5(x-1)+4=0

Answer 1

${(x - 1)}^{2} - 5(x - 1) + 4 = 0$

$(x + 1) = t$

${t}^{2} - 5t + 4 = 0$

$d = {b}^{2} - 4ac = {( - 5)}^{2} - 4 \times 1 \times 4 = 25 - 16 = 9$

$t1 = \frac{ - b - \sqrt{d} }{2a} = \frac{5 - \sqrt{9} }{2 \times 1} = 1 \\ x \\t2 = \frac{ - b + \sqrt{d} }{2a} = \frac{5 + \sqrt{9} }{2 \times 1} = 4$

$(x1 +1 ) = t1 \\ x1 = 1 - 1 \\ x1 = 0$

$x2 + 1 = t2 \\ x2 + 1 = 8 \\ x2 = 7$