[tex](2x+1-\frac{1}{1-2x} ):( 2x-\frac{4x^2}{2x-1} )[/tex]

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вопрос

To simplify the given expression, first find a common denominator for the fractions involved.

[ (2x+1-\frac{1}{1-2x}) : (2x - \frac{4x^2}{2x-1}) ]

Multiply the numerator and denominator of the first fraction by (1-2x) to get a common denominator:

[ (2x(1-2x) + (1-2x) - \frac{1(1-2x)}{1-2x}) : (2x - \frac{4x^2}{2x-1}) ]

Now simplify the first fraction:

[ (2x - 4x^2 + 1 - 2x - 1 + 2x) : (2x - \frac{4x^2}{2x-1}) ]

[ (-4x^2 + 2) : (2x - \frac{4x^2}{2x-1}) ]

Now simplify further by multiplying the second fraction by the reciprocal of the fraction inside it:

[ (-4x^2 + 2) : (2x - \frac{4x^2}{2x-1}) ]

[ (-4x^2 + 2) : (2x - \frac{4x^2(2x-1)}{1}) ]

[ (-4x^2 + 2) : (2x - 8x(2x-1)) ]

[ (-4x^2 + 2) : (2x - 16x^2 + 8x) ]

[ (-4x^2 + 2) : (2x - 16x^2 + 8x) ]

[ (-4x^2 + 2) : (-16x^2 + 10x) ]

Thus, the simplified expression is [-\frac{4x^2 - 2}{-16x^2 + 10x} ].