№ 3.6 Алгебра = № 3.6 Математика
Спростіть вираз:
1)$\frac{3x\ -\ 7y\ }{4xy} + \frac{15y\ -\ 3x\ }{4xy};$
2)$\ \frac{7a\ +\ p^3\ }{3p}\ -\ \frac{7a\ -\ 2p^3\ }{3p};$
3)$\ \frac{5a\ -\ b^4}{6b^5}\ -\ \frac{b^4\ +\ 5a\ }{6b^5};$
4)$\frac{3a\ -\ 4\ }{8a} + \frac{4a\ +\ 5\ }{8a} − \frac{1\ -\ a\ }{8a}.$
Розв'язок:
1)$\frac{3x\ -\ 7y\ }{4xy} + \frac{15y\ -\ 3x\ }{4xy} = $
$= \frac{3x\ -\ 7y\ +\ 15y\ -\ 3x\ \ }{4xy} = $
$= \frac{8y\ }{4xy} = \frac{2\ }{x};$
2)$\ \frac{7a\ +\ p^3\ }{3p}\ -\ \frac{7a\ -\ 2p^3\ }{3p}\ = $
$= \ \frac{7a\ +\ p^3\ -\ (7a\ -\ 2p^3)\ }{3p}\ =$
$= \ \frac{7a\ +\ p^3\ -\ 7a\ +\ 2p^3\ }{3p}\ =$
$= \frac{3p^3\ }{3p}\ =\ p^2;$
3)$\ \frac{5a\ -\ b^4}{6b^5}\ -\ \frac{b^4\ +\ 5a\ }{6b^5}\ =$
$= \ \frac{5a\ -\ b^4\ -\ (b^4\ +\ 5a)}{6b^5}\ =$
$= \ \frac{5a\ -\ b^4\ -\ b^4\ -\ 5a\ }{6b^5}\ =\ \frac{-2b^4\ }{6b^5}\ =$
$= -\frac{1}{3b};$
4)$\frac{3a -\ 4}{8a} + \frac{4a +\ 5 }{8a} −\frac{1 - a\ }{8a} = $
$ =\frac{3a - 4\ + \ 4a\ + 5\ - (1 - a)\ }{8a} =$
$= \frac{7a\ +\ 1 - 1 + a }{8a} = $
$=\frac{8a\ }{8a} = 1.$