Question

(x-2)×(x²-x+3)+(x²+5)×(x-3) равно 9​

Answer 1

Ответ:

3

Объяснение:

$(x-2)(x^2-x+3)+(x^2+5)(x-3)=9\\x^3-2x^2-x^2+2x+3x-6+x^3+5x-3x^2-15=9\\2x^3-6x^2+10x-30=0\;|:2\\x^3-3x^2+5x-15=0\\(x^3-3x^2)+(5x-15)=0\\x^2(x-3)+5(x-3)=0\\(x-3)(x^2+5)=0\\x-3=0\; \; \; x^2+5\neq0\\x=3$

***Примечание:

x²+5>0 для любого х∈(-∞;+∞), т.к. х²≥0 и 5>0

Answer 2

To solve the equation:

(x - 2) × (x² - x + 3) + (x² + 5) × (x - 3) = 9

First, let's simplify the expressions within each set of parentheses:

(x - 2) × (x² - x + 3) = x³ - x² + 3x - 2x² + 2x - 6 = x³ - 3x² + 5x - 6

(x² + 5) × (x - 3) = x³ - 3x² + 5x - 15

Now, substituting these simplified expressions back into the original equation:

(x³ - 3x² + 5x - 6) + (x³ - 3x² + 5x - 15) = 9

Combining like terms:

2x³ - 6x² + 10x - 21 = 9

Next, let's move all terms to one side of the equation:

2x³ - 6x² + 10x - 21 - 9 = 0

2x³ - 6x² + 10x - 30 = 0

Now, we have a cubic equation. To solve it, we can try factoring, but it may not always be possible. In this case, factoring might not yield rational roots, so let's use numerical methods or a graphing calculator to find the approximate solutions.