Question

A+(1/b)=3 ,b+(1/a)=5 , найдите (a+b)/(a-b)=?​

Answer 1

$\left \{ {{a+\frac{1}{b}=3|*b} \atop {b+\frac{1}{a}=5|*a}} \right. \left \{ {{ab+1=3b} \atop {ab+1=5a}} \right. \Rightarrow 3b=5a \Leftrightarrow a=0.6b\\\frac{a+b}{a-b}=\frac{0.6b+b}{0.6b-b}=\frac{1.6a}{-0.4b}=-4$

Ответ: -4

Answer 2

To find the value of $\dfrac{a+b}{a-b}$, we first need to solve for $a$ and $b$ in the given equations:

1. $A + \dfrac{1}{b} = 3$ (Equation 1)
2. $b + \dfrac{1}{a} = 5$ (Equation 2)

Let's solve for $a$ and $b$:

Step 1: Solve Equation 1 for $A$: $A = 3 - \dfrac{1}{b}$

Step 2: Solve Equation 2 for $b$: $b = 5 - \dfrac{1}{a}$

Step 3: Substitute the value of $b$ from Step 2 into Equation 1: $A = 3 - \dfrac{1}{5 - \dfrac{1}{a}}$

Step 4: Find the common denominator for $A$: $A = 3 - \dfrac{1}{\dfrac{5a - 1}{a}}$ $A = 3 - \dfrac{a}{5a - 1}$

Step 5: Find the common denominator for $A$: $A = \dfrac{15 - a}{5a - 1}$

Now, we have $A$ in terms of $a$. Next, we'll find the value of $a$ by setting $A$ to 0 (since $A + \dfrac{1}{b} = 3$ and we already know $A = 3 - \dfrac{1}{b}$):

$\dfrac{15 - a}{5a - 1} = 0$

To solve for $a$, we set the numerator equal to 0:

$15 - a = 0$

$a = 15$

Now that we have the value of $a$, we can find the value of $b$ by using Equation 2:

$b = 5 - \dfrac{1}{a} = 5 - \dfrac{1}{15} = \dfrac{74}{15}$

Now, we have $a = 15$ and $b = \dfrac{74}{15}$.

Finally, let's find $\dfrac{a+b}{a-b}$:

$\dfrac{a+b}{a-b} = \dfrac{15 + \dfrac{74}{15}}{15 - \dfrac{74}{15}}$

To simplify the expression, we get a common denominator:

$\dfrac{a+b}{a-b} = \dfrac{\dfrac{15 \cdot 15 + 74}{15}}{\dfrac{15 \cdot 15 - 74}{15}}$

$\dfrac{a+b}{a-b} = \dfrac{\dfrac{299}{15}}{\dfrac{176}{15}}$

Now, we can divide:

$\dfrac{a+b}{a-b} = \dfrac{299}{15} \times \dfrac{15}{176}$

$\dfrac{a+b}{a-b} = \dfrac{299}{176}$