Question

7 )7(1-х) =(2х+3)(1-х) ;8 )(3х-1)(х+3)=x(1+3x)​

Answer 1

Ответ:

7)x1=(3-1)/2=1

x2=(3+1)/2=2

8)x=3/7

Пошаговое объяснение:7)7-7x=2x-2x^2+3-3x

7-7x=3-2x^2-x

2x^2-6x+4=0

x^2-3x+2=0

D=9-4*2

D=1

x1=(3-1)/2=1

x2=(3+1)/2=2

8)(3х-1)(х+3)=x(1+3x)​

3x^2+9x-x-3=x+3x^2

9x-x-3=x

8x-x=3

7x=3

x=3/7

Answer 2

It looks like you've provided two equations. Let's solve them step by step.

Equation 1: $7(1-x) = (2x+3)(1-x)$

First, distribute the right-hand side: $7 - 7x = 2x - 2x^2 + 3 - 3x$

Combine like terms: $7 - 7x = 5x - 2x^2 + 3$

Rearrange the equation to a quadratic form: $2x^2 + 5x - 3 = 0$

Now, let's solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 2$, $b = 5$, and $c = -3$: $x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot -3}}{2 \cdot 2}$ $x = \frac{-5 \pm \sqrt{49}}{4}$ $x = \frac{-5 \pm 7}{4}$

So, the solutions for Equation 1 are: $x = \frac{1}{2}$ and $x = -3$

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Equation 2: $8(3x-1)(x+3) = x(1+3x)$

First, distribute both sides: $24x - 8 + 8x^2 + 24x = x + 3x^2$

Combine like terms: $48x - 8 + 8x^2 = 4x^2 + x$

Move all terms to one side of the equation: $8x^2 - 4x^2 + 48x - x - 8 = 0$

Simplify: $4x^2 + 47x - 8 = 0$

We have another quadratic equation. Let's solve it using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 4$, $b = 47$, and $c = -8$: $x = \frac{-47 \pm \sqrt{47^2 - 4 \cdot 4 \cdot -8}}{2 \cdot 4}$

The solutions for Equation 2 will involve more complex values. If you calculate the discriminant inside the square root, you'll find that it's a positive number. This means the solutions will be real and complex conjugates.

So, the solutions for Equation 2 are: $x \approx -5.41$ (real) and $x \approx 1.86$ (real)

Please note that the actual numerical values are rounded to two decimal places.