Question

(х-у)²+(х-3)²+|2x+y+z|=0

Answer 1

(х-у)²+(х-3)²+|2x+y+z|=0

слева сумма двух квадратов и модуля

Квадрат и модуль по определению числа всегда большие или равные 0

Значит для того чтобы сумма равнялась 0, надо чтобы каждый член равнялся 0

x-y=0 (1)

x-3=0     (2)  

2x+y-z=0  (3)

x=3 из (2)

x=y=3 из (1)

z=2x+y=6+3=9 из (3)

ответ (3 3 9)

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Answer 2

To analyze the expression and find the solution, we need to work through the equation step by step. The given equation is:

$(x - u)^2 + (x - 3)^2 + |2x + y + z| = 0$

Here, $x$, $y$, and $z$ are variables, and $u$ is a constant.

Step 1: Simplify the equation Let's start by expanding the squares and simplifying the expression:

$x^2 - 2ux + u^2 + x^2 - 6x + 9 + |2x + y + z| = 0$

Step 2: Analyze the absolute value term For the equation to be valid, the absolute value term must be non-negative. That is, $|2x + y + z| \geq 0$.

Step 3: Solve for the absolute value term To handle the absolute value, we need to consider two cases: $2x + y + z \geq 0$ and $2x + y + z < 0$.

Case 1: $2x + y + z \geq 0$ In this case, the absolute value term $|2x + y + z|$ can be replaced with $2x + y + z$. So, the equation becomes:

$x^2 - 2ux + u^2 + x^2 - 6x + 9 + (2x + y + z) = 0$

Simplifying further:

$2x^2 - 2ux + u^2 - 4x + 9 + y + z = 0$

Step 4: Combine like terms Combine the $x^2$ and $x$ terms:

$2x^2 - (2u + 4)x + u^2 + 9 + y + z = 0$

Step 5: Case 1 solution To solve for $x$ in terms of $u$, $y$, and $z$, we can use the quadratic formula:

$x = \frac{-(b) \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 2$, $b = -(2u + 4)$, and $c = u^2 + 9 + y + z$.

So, the two possible solutions for $x$ in case 1 are:

$x_1 = \frac{-(2u + 4) + \sqrt{(2u + 4)^2 - 4 \cdot 2 \cdot (u^2 + 9 + y + z)}}{2 \cdot 2}$

$x_2 = \frac{-(2u + 4) - \sqrt{(2u + 4)^2 - 4 \cdot 2 \cdot (u^2 + 9 + y + z)}}{2 \cdot 2}$

Step 6: Case 2: $2x + y + z < 0$ In this case, the absolute value term $|2x + y + z|$ becomes $-(2x + y + z)$. So, the equation becomes:

$x^2 - 2ux + u^2 + x^2 - 6x + 9 - (2x + y + z) = 0$

Simplifying further:

$2x^2 - 2ux + u^2 - 4x + 9 - y - z = 0$