. Permure ypaBHeHne: a) (t^2-8t)/(4)=0 6) (x+1)/(x-2)-(x)/(x+2)=(12)/(x^2)-4

a) To solve the equation $\frac{t^2 - 8t}{4} = 0$, we can start by multiplying both sides of the equation by 4 to eliminate the denominator:<br /><br />$t^2 - 8t = 0$<br /><br />Next, we can factor out a common factor of $t$ from the left side of the equation:<br /><br />$t(t - 8) = 0$<br /><br />Now, we have a product of two factors equal to zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. So, we can set each factor equal to zero and solve for $t$:<br /><br />$t = 0$ or $t - 8 = 0$<br /><br />Solving the second equation, we get:<br /><br />$t = 8$<br /><br />Therefore, the solutions to the equation $\frac{t^2 - 8t}{4} = 0$ are $t = 0$ and $t = 8$.<br /><br />b) To solve the equation $\frac{x+1}{x-2} - \frac{x}{x+2} = \frac{12}{x^2 - 4}$, we can start by finding a common denominator for the fractions on the left side of the equation. The common denominator is $(x-2)(x+2)$.<br /><br />Multiplying both sides of the equation by the common denominator, we get:<br /><br />$(x+1)(x+2) - x(x-2) = 12$<br /><br />Expanding and simplifying the left side of the equation, we have:<br /><br />$x^2 + 3x + 2 - x^2 + 2x = 12$<br /><br />Combining like terms, we get:<br /><br />$5x + 2 = 12$<br /><br />Subtracting 2 from both sides, we have:<br /><br />$5x = 10$<br /><br />Dividing both sides by 5, we get:<br /><br />$x = 2$<br /><br />However, we need to check if this solution is valid by substituting it back into the original equation. Substituting $x = 2$ into the equation, we get:<br /><br />$\frac{2+1}{2-2} - \frac{2}{2+2} = \frac{12}{2^2 - 4}$<br /><br />Simplifying the fractions, we have:<br /><br />$\frac{3}{0} - \frac{1}{2} = \frac{12}{0}$<br /><br />Since division by zero is undefined, the solution $x = 2$ is not valid.<br /><br />Therefore, there are no valid solutions to the equation $\frac{x+1}{x-2} - \frac{x}{x+2} = \frac{12}{x^2 - 4}$.