Question

Помогите решить примеры 1) 25-144х⁴=8х²+15        ⁻⁻⁻⁻⁻        12х²-52)1    -    а      -     4а+12-а³   ⁻⁻⁻⁻    ⁻⁻⁻⁻         ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   а-2    а²+2а+4       а³-8

Answer 1

1)(5-12x²)(5+12x²)/(12x²-5)=8x²+15
-(12x²+5)=8x²+15    12x²-5≠0⇒x≠-√5/12 x≠√5/12
8x²+5+12x²+5=0
20x²+10=0
10(2x²+1)=0-решения нет,т.к.>0при любом х
2)1/(a-2) - a/(a²+2а+4)  + (а+6)(а-2)/(а-2)(a²+2а+4)=
=1/(a-2) - a/(a²+2а+4)  + (а+6)/(a²+2а+4)=(a²+2а+4-а²+2а+а²-2а+6а-12)/(a²+2а+4)=
=(а²+8а-8)/(a²+2а+4)
Оз х²+2х+4

Answer 2

We can start by simplifying both sides of the equation and bringing all the terms to one side:

25 - 144x⁴ = 8x² + 15 - 12x² + 52

Simplifying the right side, we get:

25 - 144x⁴ = -4x² + 67

Bringing all the terms to one side:

144x⁴ + 4x² - 42 = 0

We can simplify this equation by dividing all the terms by 2:

72x⁴ + 2x² - 21 = 0

This is a quadratic equation in x², so we can use the quadratic formula to solve for x²:

x² = [-2 ± sqrt(2² - 4(72)(-21))] / (2(72))

x² = [-2 ± sqrt(5774)] / 144

x² ≈ -0.228 or x² ≈ 0.098

Since we cannot take the square root of a negative number, the only valid solution is:

x² ≈ 0.098

Taking the square root of both sides, we get:

x ≈ ±0.313

Therefore, the solutions to the equation are:

x ≈ -0.313 and x ≈ 0.313

We can start by simplifying the expression on the right side of the equation:

-а - 4а + 12 - а³ = -5а - а³ + 12

Now we can substitute this expression into the original equation:

1 - 5а - а³ + 12 = а - 2 + а² + 2а + 4

Simplifying both sides:

-а³ - а² + 4а + 9 = 0

We can factor out a negative sign:

-1(a³ + a² - 4а - 9) = 0

Now we can factor the expression inside the parentheses using grouping:

(a³ + a²) - (4а + 9) = 0

a²(a + 1) - 4(a + 1) = 0

(a + 1)(a² - 4) = 0

Solving for the roots, we get:

a + 1 = 0 or a² - 4 = 0

a = -1 or a = -2 or a = 2

However, we need to check if any of these solutions make the denominator of the expression on the left side of the original equation equal to zero:

a - 4a + 12 - a³ = (a - 2)(a² + 2a + 4)

This expression has no real roots, so none of the solutions we found make the denominator equal to zero.

Therefore, the solutions to the equation are:

a = -1, a = -2, and a = 2.