Question

Me podrían ayudar por favor

Answer 1

Respuesta:

a.

$\frac{a}{b}+\frac{ab}{c} \\a = -2 ,b=4, c= 6\\\frac{-2}{4}+\frac{(-2)(4)}{6} = -\frac{1}{2}-\frac{8}{6}= -\frac{1}{2}-\frac{4}{3}=-\frac{3+8}{6}=-\frac{11}{6}$

b.

$b^2-c^3+5+3c \\b = -3,c=-6\\(-3)^2-(-6)^3+5+3(-6) = 9-(-216)+5-18=9+216+5-18=212$

c.

$a+b+\frac{ab}{d}+3a(5b)-c\\a=10,b=-9,c=-8,d=5\\\\10+(-9)+\frac{(10)(-9)}{5}+3(10)(5(-9))-(-8) \\=10-9-\frac{90}{5}+30(-45)+8\\=10-9-18-1350+8\\=-1359$

d.

$(a)(\frac{b}{c} )(\frac{4a}{b} )(\frac{5}{4} )\\a=2,b=3,c=-4\\(2)(-\frac{3}{4} )(\frac{4(2)}{3} )(\frac{5}{4})\\=(2)(-\frac{3}{4} )(\frac{8}{3} )(\frac{5}{4})\\=-\frac{6}{4} (\frac{8}{3} )(\frac{5}{4})\\=-\frac{48}{12}(\frac{5}{4})\\=-\frac{240}{48}\\= -5$

e.

$a+b(\frac{4a}{b}(3a-d))^2+4c-10 \\a=2,b=3,c=-5,d=-3\\2+3(\frac{4(2)}{3}(3(2)-(-3)))^2+4(-5)-10\\=2+3(\frac{8}{3}(6+3))^2+4(-5)-10\\=2+3(\frac{8}{3}(9))^2+4(-5)-10\\=2+3(\frac{72}{3})^2+4(-5)-10\\=2+3(24)^2+4(-5)-10\\=2+3(576)+4(-5)-10\\=2+1728-20-10\\=1700$