Question

ajutorr dau 100 puncte pentru ex. 9​

Answer 1

$a)(x - \frac{1}{2} ) {}^{2} - ( \frac{1}{2} ) {}^{2} = (x - \frac{1}{2} - \frac{1}{2} ) \times (x - \frac{1}{2} + \frac{1}{2} ) = (x - 1) \times x = > x \times (x - 1)$

$b)(2x + \frac{1}{3} ) {}^{2} - ( \frac{1}{3} ) {}^{2} = (2x + \frac{1}{3} - \frac{1}{3} ) \times (2x + \frac{1}{3} + \frac{1}{3} ) = 2x \times (2x + \frac{2}{3} ) = 2x \times \frac{1}{3} \times (6x + 2) = 2x \times \frac{1}{3} \times 2(3x + 1) = 2 \times 2 \times \frac{1}{3} x \times (3x + 1) = \frac{4}{3} x \times (3x + 1).$

$c)( \frac{3}{4} - x) {}^{2} - \frac{1}{16 } = ( \frac{3}{4} - x - \frac{1}{4} ) \times ( \frac{3}{4} - x + \frac{1}{4} ) = ( \frac{1}{2} - x) \times (1 - x) = \frac{1}{2} \times (1 - 2x) \times (1 - x).$

$d)( \frac{x}{3} + 2y) {}^{2} - 9y {}^{2} = ( \frac{x}{3} + 2y - 3y) \times ( \frac{x}{3} + 2y + 3y) = ( \frac{x}{3} - y) \times ( \frac{x}{3} + 5y)$

$e)( \frac{x}{5} - 2y) {}^{2} - ( \frac{4x}{5} + 2y) {}^{2} =$

[

$\frac{x}{5} - 2y - ( \frac{4x}{5} + 2y)$

]

$\times$

[

$\frac{x}{5} - 2y + ( \frac{4x}{5} + 2y)$

]

$= ( \frac{x}{5} - 2y - \frac{4x}{5} - 2y) \times ( \frac{x}{5} -2y + \frac{4x}{5} + 2y) = ( - \frac{3}{5} x - 4y)x = x \times ( - \frac{3}{5}x - 4y).$

$f)(3a - \frac{1}{3} x) {}^{2} - \frac{4}{9} x = (3a) {}^{2} - 2 \times 3a \times \frac{1}{3}x + ( \frac{1}{3} x) {}^{2} - \frac{4}{9} x = 9a {}^{2} - 2ax + \frac{1}{9} x {}^{2} - \frac{4}{9} x$