Matemáticas, pregunta formulada por emixdxd333, hace 8 meses

xfavor alguien las puede hacer? :( ​

Adjuntos:

Respuestas a la pregunta

Contestado por kat388
1

a.

2x +  \frac{3}{5}  = 3 +  \frac{1}{6}

2x +  \frac{3}{5}  =  \frac{3*6}{6}  +  \frac{1}{6}

2x +  \frac{3}{5}  =   \frac{3*6 + 1}{6}

2x +  \frac{3}{5}  =  \frac{18 + 1}{6}

2x +  \frac{3}{5}  =  \frac{19}{6}

2x =  \frac{19}{6}  -  \frac{3}{5}

2x =  \frac{77}{30}

x =  \frac{ \frac{77}{30} }{2}

x =  \frac{77}{60}

-------------------------------

b.

 \frac{1}{2} x + (1 -  \frac{1}{3} ) =  \frac{1}{4}   + 1

 \frac{1}{2} x =  \frac{1}{4}  + 1 - (1 -  \frac{1}{3} )

 \frac{1}{2} x =  \frac{1}{4}  + 1 - \frac{2}{3}

\frac{1}{2} x =  \frac{1}{4}  -  \frac{2}{3} + 1

\frac{1}{2} x =  \frac{1}{4}  -  \frac{2}{3} +  \frac{1}{1}

\frac{1}{2} x =   \frac{1}{1}  + \frac{1}{4}  -  \frac{2}{3}

\frac{1}{2} x =   \frac{12}{12}  + \frac{3}{12}  -  \frac{8}{12}

 \frac{1}{2} x =  \frac{12 + 3 - 8}{12}

 \frac{1}{2} x =  \frac{15 - 8}{12}

 \frac{1}{2} x =  \frac{7}{12}

x =  \frac{7}{12} *2

x =  \frac{7}{6}

-------------------------------

c.

 \frac{2}{3} x +  \frac{1}{3}  =  \frac{4}{9} x +  \frac{1}{2}

 \frac{2}{3} x +   =  \frac{4}{9} x +  \frac{1}{2}  -  \frac{1}{3}

 \frac{2}{3} x +   =  \frac{4}{9} x +  \frac{1}{6}

 \frac{2}{3} x +   = \frac{1}{6} +   \frac{4}{9} x

 \frac{2}{3} x -  \frac{4}{9} x =  \frac{1}{6}

 \frac{2}{9} x =  \frac{1}{6}

2x =  \frac{1}{6} *9

2x =  \frac{3}{2}

x =  \frac{ \frac{3}{2} }{2}

x =  \frac{3}{4}

-------------------------------

d.

7x -  \frac{7}{10}  = 2x +  \frac{14}{5}

7x = 2x +  \frac{14}{5}  +  \frac{7}{10}

7x = 2x +  \frac{28}{10}  +  \frac{7}{10}

7x = 2x +  \frac{28 + 7}{10}

7x = 2x +  \frac{35}{10}

7x = 2x +  \frac{7}{2}

7x - 2x =  \frac{7}{2}

5x =  \frac{7}{2}

x =  \frac{ \frac{7}{2} }{5}

x =  \frac{7}{10}

-------------------------------

Espero haberte ayudado;))

Corona?


emixdxd333: cómo se dan las coronas?
Otras preguntas