РЕШИТЕ ПРАВИЛЬНО ПОЛУЧИТЕ 102 ПНК

3) 4x-13<7x-1 5x+16>9x+4

4x-7x<-1+13 5x-9x>4-16

-3x<12 -4x>-12

x>-12/3 x<12/4</p>

x>-4 x<3</p>

-4

№) x^2-8x+16=0

x1+x2=8

x1*x2=16

x1=x2=4

$(\frac{x-3}{(x-4)(x-4)}-\frac{x+5}{(x-4)(x+4)})^{-1}\frac{8(x+2)}{(x-4)(x+4)}+11=\\\\= (\frac{(x-3)(x+4)-(x+5)(x-4)}{(x-4)(x-4)(x+4)})^{-1}\frac{8(x+2)}{(x-4)(x+4)}+11=\\\\= \frac{(x-4)(x-4)(x+4)}{x^{2}-3x+4x-12-x^{2}+4x-5x+20}\cdot\frac{8(x+2)}{(x-4)(x+4)}+11=\\\\= \frac{(x-4)(x-4)(x+4)8(x+2)}{8(x-4)(x+4)}+11=(x-4)(x+2)+11=\\\\=x^{2}-4x+2x-8+11=x^{2}-2x+3$

1) $(\frac{a+2}{a-2}- \frac{a-2}{a+2}) (\frac{-(a^2-4)}{a^{2}+4})= (\frac{(a+2)(a+2)-(a-2)(a-2)}{(a-2)(a+2)})(\frac{-(a^2-4)}{a^{2}+4})=\\\\= (\frac{a^{2}+4a+4-a^{2}+4a-4}{(a-2)(a+2)})(\frac{-(a^2-4)}{a^{2}+4})=\frac{-8a(a^2-4)}{(a^{2}-4)(a^{2}+4)}=-\frac{8a}{a^{2}+4}$

2) $5\sqrt{12}+\frac{1}{4}\sqrt{48}-6\sqrt{75}=5\sqrt{4}\sqrt{3}+\frac{1}{4}\sqrt{16}\sqrt{3}-6\sqrt{25}\sqrt{3}=\\=10\sqrt{3}+\sqrt{3}-30\sqrt{30}=-19\sqrt{3}$