|
\(\Large\frac{a+\frac{a}{b}}{a-\frac{a}{b}}=\)
|
A. \(-1\) |
|---|
| B. \(\frac{1-b^2}{b}\) |
| C. \(\frac{a^2}{b^2}(b^2-1)\) |
| D. \(\frac{1}{b^2}\) |
| E. \(\frac{a^2}{b^2}(1-b^2)\) |
| F. \(\frac{b+1}{b-1}\) |
[ B-6142 - op net sinds 25.10.2024-(E)-20.10.2026 ]
Translation in E N G L I S H
|
\(\Large\frac{a+\frac{a}{b}}{a-\frac{a}{b}}=\)
|
A. \(-1\) |
|---|
| B. \(\frac{1-b^2}{b}\) |
| C. \(\frac{a^2}{b^2}(b^2-1)\) |
| D. \(\frac{1}{b^2}\) |
| E. \(\frac{a^2}{b^2}(1-b^2)\) |
| F. \(\frac{b+1}{b-1}\) |
Oplossing - Solution
1ste manier :
\(\large\frac{a+\frac{a}{b}}{a-\frac{a}{b}}=\frac{\frac{ab}{b}+\frac{a}{b}}{\frac{ab}{b}-\frac{a}{b}}=\frac{\frac{ab+a}{b}}{\frac{ab-a}{b}}=\frac{ab+a}{b}\cdot\frac{b}{ab-a}=\frac{a(1+b)}{a(b-1)}=\frac{b+1}{b-1}\)
2de manier :
\(\large\frac{a+\frac{a}{b}}{a-\frac{a}{b}}=\frac{1+\frac1b}{1-\frac1b}=\frac{\frac{b+1}{b}}{\frac{b-1}{b}}=\frac{b+1}{b}\cdot\frac{b}{b-1}=\frac{b+1}{b-1}\)
