Question

(х^2-4x+1)(x^2-4x+ 2) = 2

Answer 1

(х²-4x+1)(x²-4x+2) = 2

Пусть х²-4x+1 = t, тогда

x²-4x+ 2 = x²- 4x + 1 + 1 = t+1

t*(t+1)=2

t²+t-2=0

D = 1² - 4*(-2) = 1+8 = 9 = 3²

$t_{1} =\frac{-1+3}{2*1} = \frac{2}{2} = 1$

$t_{2} =\frac{-1-3}{2*1} = -\frac{4}{2} = -2$

Вернёмся к замене:

$\left[\begin{array}{ccc}x^{2}-4x+1=1 \\x^{2}-4x+1=-2\\\end{array}$

х²-4x+1 = 1

х²-4x = 0

х*(х-4) =0

х = 0

х-4=0

х=4

х²-4x+1 = -2

х²-4x+3 = 0

D = (-4)² - 4*3 = 16-12 = 4 = 2²

$x_{1} =\frac{4+2}{2*1} = \frac{6}{2} = 3$

$x_{2} =\frac{4-2}{2*1} = \frac{2}{2} = 1$

Ответ: x = 0, x = 1, x = 3, x = 4

Answer 2

Ответ:

Объяснение:

пусть x²-4x+1=t, тогда

t(t+1)=2

t²+t-2=0

D= 1+8=9

t1= (-1-3)/2= -2

t2= (-1+3)/2= 1

1) x²-4x+1=-2

x²-4x+3=0

D= 16-12= 4

x1= (4-2)/2= 1

x2= (4+2)/2= 3

2) x²-4x+1=1

x(x-4)=0

x3= 0

x4= 4

Ответ: x1=1, x2= 3, x3=0, x4=4