Question

Сократите дробь: 1. 3-√a/a-9 2. √a+√b/a-b

Answer 1

Ответ:

1. - 1/(√a + 3).

2. 1/(√a - √b).

Объяснение:

1. (3-√a)/(a-9) = (3-√a)/((√a)^2 - 3^2) = (3-√a)/((√a - 3)•(√a + 3)) = - (√a - 3)/((√a - 3)•(√a + 3)) = - 1/(√a + 3).

2. (√a+√b)/(a-b) = (√a+√b)/((√a)^2 - (√b)^2) = (√a+√b)/((√a - √b)•(√а + √b)) = 1/(√a - √b).

Answer 2

To simplify the given fractions, we'll rationalize the denominators. Rationalizing the denominator involves getting rid of any radicals in the denominator by multiplying both the numerator and denominator by an appropriate expression. Let's simplify the fractions:

1. $\frac{3-\sqrt{a}}{a-9}$:

To rationalize the denominator, we can multiply the numerator and denominator by $(a+9)$ (the conjugate of $(a-9)$):

$\frac{3-\sqrt{a}}{a-9} \times \frac{a+9}{a+9} = \frac{(3-\sqrt{a})(a+9)}{(a-9)(a+9)}$

Expanding the numerator:

$\frac{3a+27-\sqrt{a}a-\sqrt{a}\cdot9}{a^2-9^2} = \frac{(3a+27-a\sqrt{a}-9\sqrt{a})}{(a+9)(a-9)}$

Simplifying the numerator:

$\frac{(3a+27-a\sqrt{a}-9\sqrt{a})}{(a+9)(a-9)} = \frac{3(a+9)-\sqrt{a}(a+9)}{(a+9)(a-9)}$

Now we can cancel out the common factor $(a+9)$ from the numerator and denominator:

$\frac{3(a+9)-\sqrt{a}(a+9)}{(a+9)(a-9)} = \frac{3-\sqrt{a}}{a-9}$

So the simplified fraction is $\frac{3-\sqrt{a}}{a-9}$.

2. $\frac{\sqrt{a}+\sqrt{b}}{a-b}$:

To rationalize the denominator, we can multiply the numerator and denominator by $(\sqrt{a}-\sqrt{b})$ (the conjugate of $(a-b)$):

$\frac{\sqrt{a}+\sqrt{b}}{a-b} \times \frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \frac{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}{(a-b)(\sqrt{a}-\sqrt{b})}$