De oplossingen-
verzameling van
is
|
A.  |
B. ![https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;]\, 0,\; 2\,]](https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;]\,&space;0,\;&space;2\,]) |
C. ![https://latex.codecogs.com/png.image?\dpi{110}[\, 2,\; +\infty\, ]](https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;[\,&space;2,\;&space;+\infty\,&space;[) |
D. ![https://latex.codecogs.com/png.image?\dpi{110}]-\infty,\: -2\, ]\: \cup \: [\, 0,\; +\infty\, [](https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;]-\infty,\:&space;-2\,&space;]\:&space;\cup&space;\:&space;[\,&space;0,\;&space;+\infty\,&space;[) |
E. ![https://latex.codecogs.com/png.image?\dpi{110}]-\infty,\: 0\, ]\: \cup \: [\, 2,\; +\infty\, ]](https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;]-\infty,\:&space;0\,&space;]\:&space;\cup&space;\:&space;[\,&space;2,\;&space;+\infty\,&space;[) |
[ 5-3637 - op net sinds 3.2.15-(E)-23.11.2024]
Translation in E N G L I S H
What is the solution set of
|
A. \:&space;\cup&space;\:&space;[\,&space;2,\;&space;\infty\,&space;)) |
B. ![https://latex.codecogs.com/png.image?\dpi{110}\fn_jvn&space;]\, 0,\; 2\,]](https://latex.codecogs.com/png.image?\dpi{99}\fn_jvn&space;(\,&space;0,\;&space;2\,]) |
C. ) |
D. ![https://latex.codecogs.com/png.image?\dpi{110}]-\infty,\: -2\, ]\: \cup \: [\, 0,\; +\infty\, [](https://latex.codecogs.com/png.image?\dpi{99}\fn_jvn&space;(-\infty,\:&space;-2\,&space;]\:&space;\cup&space;\:&space;[\,&space;0,\;&space;\infty\,&space;)) |
E. ![https://latex.codecogs.com/png.image?\dpi{110}]-\infty,\: 0\, ]\: \cup \: [\, 2,\; +\infty\, ]](https://latex.codecogs.com/png.image?\dpi{99}\fn_jvn&space;(-\infty,\:&space;0\,&space;]\:&space;\cup&space;\:&space;[\,&space;2,\;&space;\infty\,&space;)) |
Oplossing - Solution
\begin{aligned}
& {\frac{4}{x}} \leq 2 \quad
\Leftrightarrow & {\frac{4}{x}} - 2 \leq 0 \quad
\Leftrightarrow & \; {\frac {4-2x}{x}} \leq 0 \\
\end{aligned}
Voor x = 2 is de teller 0, voor x = 0 is de noemer 0
Voor de andere getallen heeft de breuk hetzelfde teken als
(4 − 2x).x = − 2x² + 4x.
Dit leidt ons tot hetvolgende shcema :
x | 0 2 .
breuk | − | + 0 −
en R \ [ 0, 2 [ als oplossingenverzameling