Question

(x^2+1)^2=(x^2-1)(x^2+1)+(x+1)^2-4x+1 СРОЧНО!!!!

Answer 1

X^4+2x^2+1=x^4-1+(x+1)^2-4x+1


x^4+2x^2+1=x^4+x^2+2x+1-4x

2x^2=x^2+2x-4x

2x^2=x^2-2x

x^2+2x=0

D=b^2-4ac
D=4

x1= -2+2/2=0

x2= -2-2/2=-2

Ответ: x1=0; x2=-2;

Answer 2

Let's simplify the equation step by step:

Starting with the left-hand side (LHS):

(LHS) = (x^2 + 1)^2

Expanding the square:

(LHS) = (x^2 + 1)(x^2 + 1)

Using the FOIL method (First, Outer, Inner, Last):

(LHS) = x^2 * x^2 + x^2 * 1 + 1 * x^2 + 1 * 1

Simplifying further:

(LHS) = x^4 + x^2 + x^2 + 1

(LHS) = x^4 + 2x^2 + 1

Now let's simplify the right-hand side (RHS):

(RHS) = (x^2 - 1)(x^2 + 1) + (x + 1)^2 - 4x + 1

Expanding the first part:

(RHS) = (x^4 - 1) + (x^2 + 2x + 1) - 4x + 1

Combining like terms:

(RHS) = x^4 + x^2 + 2x - 3

Now let's compare both sides:

LHS = x^4 + 2x^2 + 1 RHS = x^4 + x^2 + 2x - 3

Since the LHS is not equal to the RHS, the original equation (x^2 + 1)^2 = (x^2 - 1)(x^2 + 1) + (x + 1)^2 - 4x + 1 is not true.

If you have any further questions, feel free to ask.