Question

Me pueden ayudar, por favor!!

Answer 1

$1-1=0$Hola, como estas? Debajo te dejo el paso a paso de cada ejercicio:

5.

$\sqrt{(1/3)^{2}-(1/5)^{2}}*(3/2) : (2/5) - (3/4)^{2} * (4/3) : (3/4)$

$\sqrt{(1/9)-(1/25)} * (15/4) - (9/16) * (16/9) =$

$\sqrt{16/225} * (15/4) - 1 =$

$(4/15) * (15/4) - 1 =$

$1-1=0$

6.

$(1/\sqrt{2})^{2}-12[((2/3-1/2):6)+(3/8(5/2-2/3))-25/36] =$

$1/2-12[((1/6):6)+(3/8(11/6))-25/36]=$

$1/2-12[1/36+11/16-25/36]=$

$1/2-12[1/48] =$

$1/2-1/4=1/4$

7.

$([\sqrt{25/36}-\sqrt{1/36} ]^{2}:(1/3))-2(5/4-1/2)^{2}=$

$([5/6-1/6]^{2}:1/3)-2(3/4)^{2}=$

$(2/3)^{2}:1/3-2(9/16)=$

$[4/9:1/3]-18/16$

$4/3-9/8=5/24$

8.

$\sqrt{(2/3):(17/27-1/3)+1/2:(1/6-1/24)} - 3/5*(7/8+13/8)=$

$\sqrt{2/3:(8/27)+1/2:(1/8)} -3/5*(5/2)$

$\sqrt{9/4+4} -3/2=$

$\sqrt{25/4} -3/2 =$

$5/2-3/2=1$

Espero te sirva, éxitos.