Question

Определи значение переменной t, при котором разность дробей 1/t−2 и 9/t+2 равна их произведению.

Answer 1

$\frac{1}{t-2} - \frac{9}{t+2} = \frac{9}{t^2-4} \\\\ \frac{t+2}{t^2-4} - \frac{9(t-2)}{t^2-4} = \frac{9}{t^2-4} \\\\ \frac{t+2-9(t-2)}{t^2-4} = \frac{9}{t^2-4} \\\\ t+2-9t+18=9 \\\\-8t=-11 \\\\ t= \frac{11}{8} =1.375$

Answer 2

Solving for the Value of t

To determine the value of the variable t at which the difference between the fractions 1/t−2 and 9/t+2 is equal to their product, we can set up the equation and solve for t.

The given equation is: 1/t−2 - 9/t+2 = (1/t−2) * (9/t+2)

Let's solve for t.

Solution:

1. Start with the given equation: 1/t−2 - 9/t+2 = (1/t−2) * (9/t+2)

2. Simplify the left-hand side of the equation: (1 - 2t) / t(t - 2) - (9 + 2t) / t(t + 2) = (9 - 18/t - 2/t + 4) / t(t - 2)

3. Combine the fractions on the left-hand side: ((1 - 2t)(t + 2) - (9 + 2t)(t - 2)) / t(t - 2)(t + 2) = (18 - 20t) / t(t - 2)

4. Set the left-hand side equal to the right-hand side: ((1 - 2t)(t + 2) - (9 + 2t)(t - 2)) / t(t - 2)(t + 2) = (18 - 20t) / t(t - 2)

5. Cross-multiply to eliminate the denominators: (1 - 2t)(t + 2) - (9 + 2t)(t - 2) = (18 - 20t)

6. Expand and simplify the equation: t + 2 - 2t^2 - 4t - 9t + 18 - 2t^2 + 4t - 18 = 18 - 20t -4t^2 - 12t + 20 = 18 - 20t -4t^2 + 8t + 2 = 0

7. Solve the quadratic equation using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a) where a = -4, b = 8, and c = 2

t = (-8 ± √(64 - 32)) / -8 t = (-8 ± √32) / -8 t = (-8 ± 4√2) / -8 t = 1 ± (1/√2)

Conclusion:

The values of t at which the difference between the fractions 1/t−2 and 9/t+2 is equal to their product are t = 1 + (1/√2) and t = 1 - (1/√2).