MaxWong

Add 1 to each equation.

$\begin{cases}xy + x + y + 1 = 24\\yz + y + z + 1= 32\\zx + z + x + 1 = 48\end{cases}$

We do this because now the left-hand side is factorable.

$\begin{cases}(x + 1)(y + 1) = 24\qquad --- (1)\\(y + 1)(z + 1) = 32\qquad --- (2)\\(z + 1)(x + 1) = 48\qquad --- (3)\end{cases}$

Now consider (1) * (2) / 3, we have

$\dfrac{(x + 1)(y + 1)^2(z + 1)}{(z + 1)(x + 1)} = \dfrac{24 \cdot 32}{48}\\ (y + 1)^2 = 16\\ y = 3\text{ or }y = -5$

When y = 3,

(x + 1)(4) = 24

x = 5

(z + 1)(5 + 1) = 48

z = 7

Therefore (x, y, z) = (5, 3, 7) is a solution.

When y = -5,

(x + 1)(-4) = 24

x = -7

(z + 1)(-7 + 1) = 48

z = -9

Therefore (x, y, z) = (-7, -5, -9) is another solution.

Solutions are (x, y, z) = (5, 3, 7) and (x, y, z) = (-7, -5, -9).

By Pythagorean theorem, $6^2 + b^2 = c^2$.

Perimeter is 15 means $6 + b + c = 15$.

Then $\begin{cases}c^2 - b^2 = 36\\b + c = 9\end{cases}$. Note that c^2 - b^2 = (c - b)(b + c) = 9(c - b). Therefore:

$9(c - b) = 36\\ c - b = 4$

Consider $2c = (b + c) + (c - b)$.

$2c = 9 + 4\\ c = \dfrac{13}2$

A similar problem is answered here: https://web2.0calc.com/questions/angles_70

Since the equation PA^2 + PB^2+PC^2=3PQ^2 + k holds true for any point P, we may let P = A, P = B, P = C and get 3 equations.

Let Q = (r, s). Then substituting P = A, B, C respectively gives: $\begin{cases}53 + 65 = 3((r - 4)^2 + (s + 1)^2) + k\\53 + 146 = 3((r + 3)^2 + (s - 1)^2) + k\\65 + 146 = 3((r - 8)^2 + (s - 6)^2) + k\end{cases}$.

Solving this system is possible, but it is very tedious work, so I will do it with Wolfram Alpha:

Check: The resulting polynomial 2x^2 + 4x + 2 is indeed the square of the binomial $(\sqrt 2 x + \sqrt 2)^2$

Simplify first.

$4x^2 + 14x +c -2x^2 - 10x = 2x^2 + 4x + c$.

If this is a square of a binomial, its discriminant is zero.

$4^2 - 4(2)(c) = 0 \\ 8c = 16\\ c = 2$

If 2u is not in the interval, then $2u \leq -10 \text{ or }2u \geq 10$. That means $u \leq -5 \text{ or }u \geq 5$.

Moreover, -20u is not in the interval, so $-20u \leq -10 \text{ or }-20u \geq 10$. That means $u \leq -\dfrac12 \text{ or }u \geq \dfrac12$.

If $u \leq -5 \text{ or }u \geq 5$, then it is true that $u \leq -\dfrac12 \text{ or }u \geq \dfrac12$. So $u \leq -\dfrac12 \text{ or }u \geq \dfrac12$ is extraneous. The range of values of u is $u \leq -5 \text{ or }u \geq 5$. Expressed in interval notation, $u \in (-\infty, -5]\cup [5, \infty)$.

You can just use a calculator, but here's a calculation trick:

The denominator looks like $n^3 - n = n(n - 1)(n + 1)$ we can perform partial fractions on $\dfrac1{n(n-1)(n+1)}$ to get $\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12 \left(\dfrac1n - \dfrac1{n + 1}\right)$. The original expression is $\displaystyle \sum_{n=2}^{10}\left(\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\left(\dfrac1n - \dfrac1{n + 1}\right)\right)=\dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1n - \dfrac1{n + 1}\right)$. Now note that each sum is a telescoping sum, so the answer is $\dfrac12 \left(1 - \dfrac1{10}\right) - \dfrac12 \left(\dfrac12 - \dfrac1{11}\right)$. You can simplify it by hand or by calculator.

The exact same problem was answered here: https://web2.0calc.com/questions/quadrilateral_18#r1

$\sqrt[3]{3- x}=-\dfrac92\\ 3 - x = \left(-\dfrac92\right)^3\\ 3 - x = -\dfrac{729}8\\ x = \dfrac{753}8$