Welke gelijkheid

is correct ? 		A.  \(\large\boldsymbol{\frac {1\,+\,ab} {1\,+\,ac} = \frac {1\,+\,b} {1\,+\,c} }\)
B.  \(\large\boldsymbol{\frac {2ab\,+\,1} {2xy\,+\,1} = \frac {ab\,+\,1} {xy\,+\,1} }\)
C.  \(\large\boldsymbol{\frac {2ab\,+\,a} {2cb\,+\,c} = \frac {a} {c} }\)
D.  \(\large\boldsymbol{\frac {a\,+\,3c} {b\,+\,3d} = \frac {a\,+\,c} {b\,+\,d} }\)
E.  \(\large\boldsymbol{\frac {a\,+\,2bc} {x\,+\,2yc} = \frac {a\,+\,2b} {x\,+\,2y} }\)
A    B    C    D    E

[ 2,3-1406 - op net sinds 15.1.14-(E)-15.7.2024 ]

Translation in   E N G L I S H

Which
equality
is correct 		A.   \(\boldsymbol{\frac {1\,+\,ab} {1\,+\,ac} = \frac {1\,+\,b} {1\,+\,c} }\)
B.   \(\boldsymbol{\frac {2ab\,+\,1} {2xy\,+\,1} = \frac {ab\,+\,1} {xy\,+\,1} }\)
C.   \(\boldsymbol{\frac {2ab\,+\,a} {2cb\,+\,c} = \frac {a} {c} }\)
D.   \(\boldsymbol{\frac {a\,+\,3c} {b\,+\,3d} = \frac {a\,+\,c} {b\,+\,d} }\)
E.   \(\boldsymbol{\frac {a\,+\,2bc} {x\,+\,2yc} = \frac {a\,+\,2b} {x\,+\,2y} }\)

Oplossing - Solution

1ste manier :
\(\frac {2ab+a} {2cb+c}= \frac ac \) want het product van de uitersten is gelijk aan het product van de middelsten :
(2ab + a).c = 2abc + ac
(2cb + c).a = 2abc + ac
2de manier :
\(\frac {2ab+a} {2cb+c}= \frac{a(2b+1)}{c(2b+1) = }\frac ac \)
gricha