Question

3)x²+11x+30 ________ x²-36 4)Розв'яжіть рівняння а)x²-10x²+9=0; b)2(x+3)²-5(x+3)+2=0; c) 2c x __ + __=0 x+1 x-1 d)x-3√x-4=0

Answer 1

Ответ:

4) х^2-10х^2+9=0

-9х^2+9=0 ( умножаем на -1)

9х^2 - 9=0

9х^2=9

х=+-1

.

2(х+3)^2-5(х+3)+2=0

2(х^2+6х+9)-5х+15+2=0

2х^2+ 12х+18-5х+15+2=0

2х^2+7х+35=0

D= 49-280=-231

нет решений

Answer 2

1. To simplify the rational expression, we factor both the numerator and denominator:

$x^2 + 11x + 30 = (x + 5)(x + 6)$

$x^2 - 36 = (x + 6)(x - 6)$

Now the rational expression becomes:

$\frac{x^2 + 11x + 30}{x^2 - 36} = \frac{(x + 5)(x + 6)}{(x + 6)(x - 6)}$

Since (x + 6) appears in both the numerator and denominator, we can cancel it out:

$\frac{x^2 + 11x + 30}{x^2 - 36} = \frac{x + 5}{x - 6}$

2. Let's solve each quadratic equation separately:

a) $x^2 - 10x^2 + 9 = 0$

Combine like terms: $-9x^2 + 9 = 0$

Subtract 9 from both sides: $-9x^2 = -9$

Divide by -9: $x^2 = 1$

Take the square root of both sides: $x = \pm 1$

So the solutions are $x = 1$ and $x = -1$.

b) $2(x + 3)^2 - 5(x + 3) + 2 = 0$

Let $y = x + 3$, then the equation becomes: $2y^2 - 5y + 2 = 0$

Factor the quadratic: $(2y - 1)(y - 2) = 0$

Set each factor to zero and solve for y:

$2y - 1 = 0 \Rightarrow 2y = 1 \Rightarrow y = \frac{1}{2}$

$y - 2 = 0 \Rightarrow y = 2$

Now substitute back $y = x + 3$:

$x + 3 = \frac{1}{2} \Rightarrow x = \frac{1}{2} - 3 \Rightarrow x = -\frac{5}{2}$

$x + 3 = 2 \Rightarrow x = 2 - 3 \Rightarrow x = -1$

So the solutions are $x = -\frac{5}{2}$ and $x = -1$.

c) $2cx \frac{1}{x + 1} + \frac{1}{x - 1} = 0$

To solve this, we need to find a common denominator. The common denominator is $x^2 - 1$:

$\frac{2cx(x - 1)}{(x + 1)(x - 1)} + \frac{(x + 1)}{(x + 1)(x - 1)} = 0$

Combine the fractions:

$\frac{2cx(x - 1) + (x + 1)}{(x + 1)(x - 1)} = 0$

Now we have:

$2cx^2 - 2cx + x + 1 = 0$

Rearrange and combine like terms:

$2cx^2 + (1 - 2c)x + 1 = 0$

Now, this is a quadratic equation in terms of $x$. For the given equation to be true, the coefficients of the quadratic must be zero:

$2c = 0$ (coefficient of $x^2$)

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