Question

cuanto da (1+2)3-4
(2-1/2)(3-2/3)+4(1 a la dos)
(2(1/2)+3(2/3))(4(1/4)+2 a la dos)
2(1/2)+6(2 a la dos+3 a la dos)-4(4 a la dos)

Answer 1

$* Primer \ c\'alculo:$

$(1+2).3-4 = \\ 3 . 3 - 4 = \\ 9 - 4 = \\ 5$


$* Segundo \ c\'alculo:$

$( \frac{2}{1} - \frac{1}{2}) . ( \frac{3}{1} - \frac{2}{3}) + 4 . 1^{2} =$

$\frac{(2:1.2)-(2:2.1)}{2} . \frac{(3:1.3)-(3:3.2)}{3} + 4 . 1^{2} =$

$\frac{4-1}{2} . \frac{9-2}{3} + 4 . 1^{2} =$

$\frac{3}{2} . \frac{7}{3} + 4 . 1^{2} =$

$\frac{3}{2} . \frac{7}{3} + 4 . 1 =$

$\frac{3.7}{2.3} + 4 =$

$\frac{21}{6} + 4 =$

$\frac{21}{6} + \frac{4}{1} =$

$\frac{(6:6.21)+(6:1.4)}{6} =$

$\frac{21+24}{6} =$

$\frac{45}{6} = \frac{15}{2}$


$* Tercer \ c\'alculo:$

$( 2 . (\frac{1}{2}) + 3 . ( \frac{2}{3}) ) . (4 . ( \frac{1}{4}) + 2^{2}) =$

$(2.( \frac{1}{2}) + 3 . (\frac{2}{3}) ) . ( 4 . ( \frac{1}{4} + 4 ) =$

$( \frac{2}{1} . \frac{1}{2} + \frac{3}{1} . \frac{2}{3}) . ( \frac{4}{1} . \frac{1}{4} + 4) =$

$( \frac{2.1}{1.2} + \frac{3.2}{1.3} ) . ( \frac{4.1}{1.4} + 4 ) =$

$( \frac{2}{2} + \frac{6}{3}) . ( \frac{4}{4} + \frac{4}{1}) =$

$\frac{(6:2.2)+(6:3.6)}{6} . \frac{(4:4.4)+(4:1.4)}{4} =$

$\frac{6+12}{6} . \frac{4+16}{4} =$

$\frac{18}{6} . \frac{20}{4} =$

$\frac{18.20}{6.4} =$

$\frac{360}{24} = \frac{15}{1} = 15$


$* Cuarto\ c\'alculo:$

$\frac{2}{1} . \frac{1}{2} + 6 . ( 2^{2} + 3^{2} ) - 4 . (4^{2}) =$

$\frac{2}{1} . \frac{1}{2} + 6 . (4+9) - 4 . 16 =$

$\frac{2}{1} . \frac{1}{2} + 6 . 13 - 4 . 16 =$

$\frac{2.1}{1.2} + 78 - 64 =$

$\frac{2}{2} + 78 - 64 =$

$\frac{1}{1} + 78 - 64 =$

$1 + 78 - 64 =$

$79 - 64 =$

$15$