Question

пожайлуста с полным решением, кто может прошу​

Answer 1

Ответ:

1)

$\frac{4^{-3}* 2^{-5} }{16*8^{-4} } = \frac{(2^{2}) ^{-3} *2^{-5}} {2^4*(2^3)^{-4}} = \frac{2^{-6}*2^{-5}}{2^4*2^{-12}} = \frac{2^{-11}}{2^{-8}} = 2^{-3} = \frac{1}{8}$

2)

$\frac{7a-ab}{14a} = \frac{7-b}{14}$

3)

$\frac{6}{\sqrt{3} } = \frac{2*3}{\sqrt{3} } = \frac{2*\sqrt{3}*\sqrt{3} }{\sqrt{3} } = 2\sqrt{3}$

4)

$y=\frac{k}{x}\\ A (-8; 0.3)\\0.3 = \frac{k}{-8} \\k = -2.4$

5)

$x^{2} -2x-3 = 0\\D = 4+4*3 = 4+12=16=4^2\\x1 = (2+4):2 = 3\\x2= (2-4):2 = -1$

6)

$\sqrt{21}-\sqrt{3} = \sqrt{7*3} - \sqrt{3} = \sqrt{7} *\sqrt{3} - \sqrt{3} = \sqrt{3}(\sqrt{7}-1)$

7)

$\frac{a+4}{a^2-6a+9} : \frac{a^2-16}{a-3} - \frac{2}{a-4} = \frac{a+4}{(a-3)^2}*\frac{a-3}{(a-4)(a+4)} - \frac{2}{a-4} = \frac{1}{(a-3)(a-4)} - \frac{2}{a-4} = \frac{1}{(a-3)(a-4)} - \frac{2(a-3)}{(a-4)(a-3)} = \frac{1-2a+6}{(a-3)(a-4)} = \frac{7-a}{(a-4)(a-3)}$