Question

(1/а^2-ав -3в^2/а^4-ав^3 -в/а^3+а^2в+ав^2)*(в+а^2/а+в)

Answer 1

$1)\frac{1}{a^{2}-ab} -\frac{3b^{2}}{a^{4}-ab^{3}} -\frac{b}{a^{3}+a^{2}b+ab^{2}}=\frac{1}{a(a-b)}-\frac{3b^{2}}{a(a^{3}-b^{3})}-\frac{b}{a(a^{2}+ab+b^{2})}=\frac{1}{a(a-b)}-\frac{3b^{2}}{a(a-b)(a^{2}+ab+b^{2})}-\frac{b}{a(a^{2}+ab+b^{2})}=\frac{a^{2}+ab+b^{2}-3b^{2}-ab+b^{2}}{a(a-b)(a^{2}+ab+b^{2})} =\frac{a^{2}-b^{2}}{a(a-b)(a^{2}+ab+b^{2})}= \frac{a+b}{a(a^{2}+ab+b^{2})}$

$2)\frac{a+b}{a(a^{2}+ab+b^{2})}*(b+\frac{a^{2}}{a+b})=\frac{a+b}{a(a^{2}+ab+b^{2})} *\frac{ab+b^{2}+a^{2}}{a+b}=\frac{1}{a}$

Answer 2

To simplify the given expression, let's first factorize the denominators and then proceed with the simplification. Here's the step-by-step process:

Step 1: Factorize the denominators. 1/(а^2 - аv - 3в^2) = 1/[(а - в)(а + 3в)] -3в^2/(а^4 - аv^3 - в/а^3 + а^2в + ав^2) = -3в^2/[(а^2 - в)(а^2 + в)(а - в)(а + в)] в/(а^3 + а^2в + ав^2) = в/[а(а^2 + в)(а + в)]

Step 2: Now, let's rewrite the expression with the factored denominators: [1/((а - в)(а + 3в))] * [-3в^2/((а^2 - в)(а^2 + в)(а - в)(а + в))] * [в/(а(а^2 + в)(а + в))] * [(в + а^2)/(а + в)]

Step 3: Next, let's look for common factors that can cancel out: -3в^2 and в in the second and third terms can cancel out, leaving us with: [1/((а - в)(а + 3в))] * [1/((а^2 - в)(а^2 + в)(а - в))] * [1/(а(а + в))] * [(в + а^2)]

Step 4: Now, let's multiply the fractions together: The denominators are already factored, so we just need to multiply the numerators: [1 * 1 * 1 * (в + а^2)] / [(а - в)(а + 3в)(а^2 - в)(а^2 + в)(а - в) * а(а + в)]

Step 5: Simplify further: The (а - в) terms in the numerator and denominator will cancel out, leaving us with: (в + а^2) / [(а + 3в)(а^2 - в)(а^2 + в)(а + в) * а]

So, the simplified expression is: (в + а^2) / [(а + 3в)(а^2 - в)(а^2 + в)(а + в) * а]