============== А1 ==============
![\frac{16x^6(a+2)^4}{12x^2(a+2)^5} = \frac{16}{12} \cdot x^{6-2}\cdot (a+2)^{4-5}= \frac{4}{3}\cdot x^ 4\cdot(a+2)^{-1}= \frac{4x^4}{3(a+2)}](https://tex.z-dn.net/?f=+%5Cfrac%7B16x%5E6%28a%2B2%29%5E4%7D%7B12x%5E2%28a%2B2%29%5E5%7D+%3D+%5Cfrac%7B16%7D%7B12%7D+%5Ccdot+x%5E%7B6-2%7D%5Ccdot+%28a%2B2%29%5E%7B4-5%7D%3D+%5Cfrac%7B4%7D%7B3%7D%5Ccdot+x%5E+4%5Ccdot%28a%2B2%29%5E%7B-1%7D%3D+%5Cfrac%7B4x%5E4%7D%7B3%28a%2B2%29%7D+)
Ответ: 1
============== А2 ==============
![\frac{8x}{3x-3y}- \frac{2x+6y}{3x-3y} = \frac{8x-(2x+6y)}{3x-3y} =\frac{8x-2x-6y}{3x-3y} =\frac{6x-6y}{3x-3y} =\frac{6(x-y)}{3(x-y)} =2](https://tex.z-dn.net/?f=+%5Cfrac%7B8x%7D%7B3x-3y%7D-+%5Cfrac%7B2x%2B6y%7D%7B3x-3y%7D++%3D+%5Cfrac%7B8x-%282x%2B6y%29%7D%7B3x-3y%7D+%3D%5Cfrac%7B8x-2x-6y%7D%7B3x-3y%7D+%3D%5Cfrac%7B6x-6y%7D%7B3x-3y%7D+%3D%5Cfrac%7B6%28x-y%29%7D%7B3%28x-y%29%7D+%3D2)
Ответ: 2
============== А3 ==============
![(2 \sqrt{5}-1 )(2 \sqrt{5}+1 )=(2 \sqrt{5})^2-1^2=4\cdot5-1=20-1=19](https://tex.z-dn.net/?f=%282+%5Csqrt%7B5%7D-1+%29%282+%5Csqrt%7B5%7D%2B1+%29%3D%282+%5Csqrt%7B5%7D%29%5E2-1%5E2%3D4%5Ccdot5-1%3D20-1%3D19)
Ответ: 2
============== А4 ==============
Оба утверждения верные.
Ответ: 3
============== B1 ==============
![\sqrt{54}\cdot \sqrt{6} = \sqrt{54\cdot 6} = \sqrt{324} =18](https://tex.z-dn.net/?f=+%5Csqrt%7B54%7D%5Ccdot+%5Csqrt%7B6%7D++%3D+%5Csqrt%7B54%5Ccdot+6%7D++%3D+%5Csqrt%7B324%7D+%3D18)
1)x-2√xy+y+x+2√xy+y=2x+2y
2)4a+12√ab+9b+4a-6√ab+6√ab-9b=12√ab
<span>1)
(5x + 4)/ (x - 3) < 4
</span>(5x + 4)/ (x - 3) - <span>(4x-12)/ (x - 3) < 0</span>
<span>(x + 16)/ (x - 3)<span>< 0
</span></span>метод интервалов
<span>(x + 16) = 0 при х=-16
</span>(x - 3)= 0<span><span><span> при х=3
ответ х є (-16;3)
</span>
</span>2) (3x - 15)/( x^2 +5x - 14) >= 0</span>
метод интервалов
<span>(3x -15) = 0 при х=5
</span>( x^2 +5x - 14) = 0 <span>при х=-7 и при </span>х=2
<span>ответ х є (-7;2) U [5;+беск)</span>