реши
To solve the given problem, let's break it down step by step.
Expression Analysis
The mathematical expression to solve is:
\[ \left(\frac{9a^2 - 1}{16b^2}\right) \div \left(\frac{3a - 1}{4b}\right) \text{ при } a = \frac{2}{3} \text{ и } b = -\frac{1}{12} \]
This is a complex fraction that involves a division of two fractions. The division of fractions can be rewritten as multiplication by the reciprocal:
\[ \frac{9a^2 - 1}{16b^2} \times \frac{4b}{3a - 1} \]
Substitute Values for \(a\) and \(b\):
Given: \[ a = \frac{2}{3} \] \[ b = -\frac{1}{12} \]
First, calculate each part of the expression separately.
Step-by-Step Calculation
- Calculate \(9a^2 - 1\):
\[ a^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \] \[ 9a^2 = 9 \times \frac{4}{9} = 4 \] \[ 9a^2 - 1 = 4 - 1 = 3 \]
- Calculate \(16b^2\):
\[ b^2 = \left(-\frac{1}{12}\right)^2 = \frac{1}{144} \] \[ 16b^2 = 16 \times \frac{1}{144} = \frac{16}{144} = \frac{1}{9} \]
- Calculate \(\frac{9a^2 - 1}{16b^2}\):
\[ \frac{9a^2 - 1}{16b^2} = \frac{3}{\frac{1}{9}} = 3 \times 9 = 27 \]
- Calculate \(3a - 1\):
\[ 3a = 3 \times \frac{2}{3} = 2 \] \[ 3a - 1 = 2 - 1 = 1 \]
- Calculate \(\frac{4b}{3a - 1}\):
\[ 4b = 4 \times \left(-\frac{1}{12}\right) = -\frac{4}{12} = -\frac{1}{3} \] \[ \frac{4b}{3a - 1} = \frac{-\frac{1}{3}}{1} = -\frac{1}{3} \]
- Final Step \((27) \times \left(-\frac{1}{3}\right)\):
\[ 27 \times \left(-\frac{1}{3}\right) = -9 \]
Solve the System of Equations
Now solve the system of linear equations:
\[ \begin{cases} 3x + 2y = 8 \\ 4x - y = 7 \end{cases} \]
- Express \(y\) from the second equation:
\[ 4x - y = 7 \implies y = 4x - 7 \]
- Substitute \(y\) into the first equation:
\[ 3x + 2(4x - 7) = 8 \] \[ 3x + 8x - 14 = 8 \] \[ 11x - 14 = 8 \] \[ 11x = 22 \] \[ x = 2 \]
- Find \(y\):
\[ y = 4x - 7 = 4(2) - 7 = 8 - 7 = 1 \]
Conclusion
Thus, the solution of the system of equations is:
\[ x = 2, \quad y = 1 \]
Therefore, \(x + y = 2 + 1 = 3\).
The final answer is \(-9\) for the expression involving \(a\) and \(b\), and \(x + y = 3\) for the system of equations.
