Question

3х(квадрат2)+(х+1) (х-5)+4=х(квадрат2)+(3х-1)(х-3)

Answer 1

$3x^2+(x+1)(x-5)+4=x^2+(3x-1)(x-3)\\3x^2+x^2-5x+x-5+4=x^2+3x^2-9x-x+3\\3x^2+x^2-5x+x-x^2-3x^2+9x+x=3+5-4\\6x=4\ /:6\\x=\frac{2}{3}$

Answer 2

To solve the equation, let's first simplify both sides:

Left-hand side (LHS): 3x² + (x + 1)(x - 5) + 4

Right-hand side (RHS): x² + (3x - 1)(x - 3)

Now, expand the expressions:

LHS: 3x² + (x² - 5x + x - 5) + 4 3x² + (x² - 4x - 5) + 4 3x² + x² - 4x - 1

RHS: x² + (3x² - 9x - x + 3) x² + (3x² - 10x + 3)

Now, combine like terms:

LHS: 3x² + x² - 4x - 1 4x² - 4x - 1

RHS: x² + 3x² - 10x + 3 4x² - 10x + 3

Now, set the LHS equal to the RHS and solve for x:

4x² - 4x - 1 = 4x² - 10x + 3

Subtract 4x² from both sides:

-4x - 1 = -10x + 3

Add 10x to both sides:

6x - 1 = 3

Add 1 to both sides:

6x = 4

Finally, divide by 6 to isolate x:

x = 4/6

Simplify the fraction:

x = 2/3

So the solution to the equation is x = 2/3.