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Solving a Russian Expression
I need to analyze a mathematical expression in Russian. The image shows the equation, and I’ll need to explain it carefully, including domain restrictions and simplifications.
Starting with the expression E = (4a - 24a/(4a-3) + 3) ÷ [((3a+6)^2 - 9(a^2+1))/(12a+9)^2], I’ll simplify the numerator and denominator carefully. I find restrictions such as a must not equal -3/4 and check if the fraction simplifies correctly to avoid division by zero. I want to ensure clarity throughout!Testing algebraic factors
I'm considering whether the expression could simplify to \( \frac{1}{4a+3} \). Maybe the first parenthesis should be \( 4a - \frac{24a}{4a-3}+3 \). I think 16a² - 24a - 9 could factor as \( (4a+?)(4a-?) \).
Let me test integer factors with constants of 9: \( (4a+3)(4a-3) = 16a² - 9\), but it lacks the -24a term. Hmm, \( (4a-?)(4a+?) \) doesn't seem to work either.Evaluating algebraic expression
I’m working with the denominator, which is \((12a + 9)^2 = (3(4a + 3))² = 9(4a + 3)²\). That means I can simplify to \(1/(4a + 3)\). I also think about whether I should have a \(\div 3\) instead of a plus 3, but it seems there’s definitely a plus inside the parentheses. It looks like I need to clarify the original expression and its domain where \(4a - 3 \neq 0\) and \(12a + 9 \neq 0\).
Краткий ответ: \[ \boxed{\frac{(4a+3)(16a^2-24a-9)}{4a-3}}, \qquad a\ne \pm\frac34. \]
Решение
Сначала упростим дробь, на которую делят:
\[ (3a+6)^2-9(a^2+1) =9(a+2)^2-9(a^2+1) \]
\[ =9\bigl(a^2+4a+4-a^2-1\bigr) =9(4a+3). \]
Знаменатель:
\[ (12a+9)^2=\bigl(3(4a+3)\bigr)^2=9(4a+3)^2. \]
Значит,
\[ \frac{(3a+6)^2-9(a^2+1)}{(12a+9)^2} = \frac{9(4a+3)}{9(4a+3)^2} = \frac1{4a+3}. \]
Теперь выражение имеет вид:
\[ \left(4a-\frac{24a}{4a-3}+3\right)\div\frac1{4a+3}. \]
Деление на дробь заменяем умножением на обратную:
\[ \left(4a+3-\frac{24a}{4a-3}\right)(4a+3). \]
Приведём выражение в скобках к общему знаменателю:
\[ 4a+3-\frac{24a}{4a-3} = \frac{(4a+3)(4a-3)-24a}{4a-3} \]
\[ = \frac{16a^2-9-24a}{4a-3} = \frac{16a^2-24a-9}{4a-3}. \]
Следовательно,
\[ \boxed{ \frac{(16a^2-24a-9)(4a+3)}{4a-3} }. \]
Ограничения: \(4a-3\ne0\) и \(12a+9\ne0\), поэтому
\[ a\ne\frac34,\qquad a\ne-\frac34. \]
Проверка при \(a=1\): исходное выражение и полученная формула дают одинаковый результат \(-119\).
