Question

1)9-4a²:4a²-12a+9 при a=2/3 2)(2x:x+y-2x²:x²+2xy+y²)*(1+2y:x-y) при x=-1,y=-1/2 3)(x:x-y+2xy:x²-2xy+y²)*(2x:x+y-1) при x=-2,y=-1

Answer 1

9-4*4/9:4*4/9-12*2/3+9=9-1*4/9*4/9-12*2/3+9=

=9-1*4/9*4/9-38/3+9=9-16/81-38/3+9=18-1042/81=

=416/81

Answer 2

Let's solve each expression step by step:

1. $9 - 4a^2 : 4a^2 - 12a + 9$ when $a = \frac{2}{3}$

First, substitute $a$ with $\frac{2}{3}$ and simplify the expression:

$9 - 4\left(\frac{2}{3}\right)^2 : 4\left(\frac{2}{3}\right)^2 - 12\left(\frac{2}{3}\right) + 9$

$9 - 4\left(\frac{4}{9}\right) : 4\left(\frac{4}{9}\right) - \frac{24}{3} + 9$

$9 - \frac{16}{9} : \frac{16}{9} - 8 + 9$

Now, perform the division and continue simplifying:

$9 - \frac{16}{9} \times \frac{9}{16} - 8 + 9$

$9 - 1 - 8 + 9$

$9 - 1 - 8 + 9 = 9$

So, the result is $9$.

2. $\left(\frac{2x}{x+y} - \frac{2x^2}{x^2+2xy+y^2}\right) \times \left(1 + \frac{2y}{x-y}\right)$ when $x = -1, y = -\frac{1}{2}$

First, substitute $x$ and $y$ with their respective values:

$\left(\frac{2(-1)}{-1 + (-\frac{1}{2})} - \frac{2(-1)^2}{(-1)^2 + 2(-1)(-\frac{1}{2}) + (-\frac{1}{2})^2}\right) \times \left(1 + \frac{2(-\frac{1}{2})}{-1 - (-\frac{1}{2})}\right)$

$\left(\frac{-2}{-\frac{3}{2}} - \frac{-2}{1 + 1 + \frac{1}{4}}\right) \times \left(1 + \frac{-1}{\frac{1}{2}}\right)$

Now, continue simplifying the expression:

$\left(\frac{-2}{-\frac{3}{2}} - \frac{-2}{1 + 1 + \frac{1}{4}}\right) \times \left(1 + \frac{-1}{\frac{1}{2}}\right)$