y²-2y+1 y=101;-11; 0,6

To solve the equation \( y^2 - 2y + 1 = 0 \), we can first recognize that this is a quadratic equation.

Let's factor the expression:

\[ y^2 - 2y + 1 = (y - 1)^2 = 0 \]

Setting the factor equal to zero gives:

\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

Now regarding the values \( y = 101, -11, 0.6 \) that you provided, they do not satisfy the equation \( y^2 - 2y + 1 = 0 \), since the only solution derived from the factoring is \( y = 1 \).

To confirm, we can substitute each value into the original equation:

1. For \( y = 101 \): \[ 101^2 - 2(101) + 1 = 10201 - 202 + 1 = 10000 \quad (\text{not } 0) \]

2. For \( y = -11 \): \[ (-11)^2 - 2(-11) + 1 = 121 + 22 + 1 = 144 \quad (\text{not } 0) \]

3. For \( y = 0.6 \): \[ (0.6)^2 - 2(0.6) + 1 = 0.36 - 1.2 + 1 = 0.16 \quad (\text{not } 0) \]

Thus, the only solution to the equation is \( y = 1 \).