Question

1) 2root(5-2X)-2root(X+17)=0 2) X^2-2-(X^2+8)/(X^2+2)=0 3) 2/(X^2-X-12)+6/(X^2+4X+3)=1/(X+3)

Answer 1

1)ОДЗ 5-2x≥0⇒x≤2,5 U x+17≥0⇒x≥-17
x∈[-17;2,5]
2√(5-2x)=2√(x+17)
5-2x=x+17
x+2x=5-17
3x=-12
x=-4
2)x²-2=(x²+8)/(x²+2)
x^4-4-x²-8=0
x^4-x^2-12=0
x²=a
a²-a-12=0
a1+a2=1 U a1*a2=-12
a1=-3⇒x²=-3 нет решения
a2=4⇒x²=4⇒x=+-2
3)2/(x-4)(x+3) +6/(x+1)(x+3)=1/(x+3)
x≠4  x≠-3    x≠-1
2(x+1)+6(x-4)=(x+1)(x-4)
x²-4x+x-4-2x-2-6x+24=0
x²-11x+18=0
x1+x2=11 U x1*x2=18
x1=2  x2=9

Answer 2

Let's solve each of these equations step by step:

1. $2\sqrt{5 - 2X} - 2\sqrt{X + 17} = 0$

First, move one of the square roots to the other side of the equation:

$2\sqrt{5 - 2X} = 2\sqrt{X + 17}$

Now, divide both sides by 2:

$\sqrt{5 - 2X} = \sqrt{X + 17}$

Now, square both sides of the equation to eliminate the square roots:

$5 - 2X = X + 17$

Now, isolate the variable $X$:

$-2X - X = 17 - 5$

$-3X = 12$

$X = -4$

So, the solution to the equation is $X = -4$.

2. $X^2 - 2 - \frac{X^2 + 8}{X^2 + 2} = 0$

First, find a common denominator for the fraction:

$X^2 - 2 - \frac{X^2 + 8}{X^2 + 2} = 0$

Multiply the first term by $\frac{X^2 + 2}{X^2 + 2}$ to get a common denominator:

$X^2(X^2 + 2) - 2(X^2 + 2) - (X^2 + 8) = 0$

Expand and simplify:

$X^4 + 2X^2 - 2X^2 - 4 - X^2 - 8 = 0$

Combine like terms:

$X^4 - 12 = 0$

Now, add 12 to both sides:

$X^4 = 12$

Take the fourth root of both sides:

$X = \pm\sqrt[4]{12}$

There are two real solutions:

$X = \sqrt[4]{12}$ and $X = -\sqrt[4]{12}$.

3. $\frac{2}{X^2 - X - 12} + \frac{6}{X^2 + 4X + 3} = \frac{1}{X + 3}$

First, factor the denominators of the fractions on the left side:

$\frac{2}{(X - 4)(X + 3)} + \frac{6}{(X + 1)(X + 3)} = \frac{1}{X + 3}$

Now, find a common denominator for the fractions on the left side, which is $(X - 4)(X + 1)(X + 3)$:

$\frac{2(X + 1)}{(X - 4)(X + 1)(X + 3)} + \frac{6(X - 4)}{(X - 4)(X + 1)(X + 3)} = \frac{(X - 4)}{(X - 4)(X + 1)(X + 3)}$