Given x2 + 2XY + y2-6x-6y + 9 = 0, find the value of X + y

Given x2 + 2XY + y2-6x-6y + 9 = 0, find the value of X + y

x" + 2xy + y" - 6x - 6y + 9 = 0
( x + y )" - 6( x + y ) + 3" = 0
( x + y - 3 )" = 0
So that's x + y = 3

Given x + 4Y + 3Z = 26, find the minimum value of X & # 178; + Y & # 178; + 4Y & # 178
And the general solution is explained

X + 4Y + 3Z = 26 is a face 1, X & # 178; + Y & # 178; + 4Z & # 178; = x & # 178; + Y & # 178; + (2Z) &# 178;
If Z1 = Z, then x + 4Y + 1.5z = 26 is face 2. The minimum value is the minimum value from the point on face 2 to this face, that is, the value from (0,0) to x + 4Y + 1.5z = 26. The result is (1 + 4 ^ 2 + 1.5 ^ 2) under 26 / root sign

We know the equation (9x Square-1) square + (3K-1) x = 0 (1) about X. if the equation is a quadratic equation of one variable, we can find the value range of K
(2) If the equation is a linear equation of one variable, find the value of K

Hello!
Wrong title. It should be
(9k² - 1)x² + (3k-1)x = 0
(1) The equation is a quadratic equation of one variable
∴ 9k² - 1 ≠ 0
k ≠ ± 1/3
(2) The equation is a linear equation of one variable
9K & # 178; - 1 = 0 and 3K-1 ≠ 0
∴ k = - 1/3