Question

(xx+1)(x+2)(x+4)=40 как решить?

Answer 1

[(x+1)(x+5)]*[(x+2)(x+4)]=40

(x²+6x+5)(x²+6x+8)=40

x²+6x+5=a

a(a+3)-40=0

a²+3a-40=0

a1+a2=-3 U a1*a2=-40

a1=-8⇒x²+6x+5=-8

x²+6x+13=0

D=36-52=-16<0 нет решения

a2=5⇒x²+6x+5=5

x²+6x=0

x(x+6)=0

x=0 U x=-6

Answer 2

To solve the equation $(xx+1)(x+2)(x+4) = 40$, we need to find the values of $x$ that satisfy this equation. Let's proceed step-by-step:

Step 1: Expand the left-hand side of the equation: $(xx+1)(x+2)(x+4) = (x^2+1)(x+2)(x+4)$

Step 2: Use the distributive property to further expand the equation: $(x^2+1)(x+2)(x+4) = (x^2+1)(x^2+6x+8)$

Step 3: Expand the expression inside the parentheses: $(x^2+1)(x^2+6x+8) = x^4 + 6x^3 + 8x^2 + x^2 + 6x + 8$

Step 4: Simplify the equation by combining like terms: $x^4 + 6x^3 + 9x^2 + 6x + 8 = 40$

Step 5: Move all terms to one side of the equation to set it equal to zero: $x^4 + 6x^3 + 9x^2 + 6x + 8 - 40 = 0$

Step 6: Combine constants on the left side: $x^4 + 6x^3 + 9x^2 + 6x - 32 = 0$

At this point, we have a quartic equation in $x$ (a polynomial equation of degree 4). Solving quartic equations can be quite involved and may not have a simple algebraic solution. We can use numerical methods or calculators to find approximate solutions.

One way to proceed is to use numerical methods like the Newton-Raphson method or graphing calculators to find approximate solutions for $x$. Alternatively, you can use computational software like Wolfram Alpha to solve the equation for you.

If you need specific solutions, please let me know, and I can help you with numerical approximations or any other information you might need.