Wiki-Quellcode von Lösung Rechnen mit Potenzen
Verstecke letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
 |
1.1 | 1 | 1.a) {{formula}}3a^2 + 5b^3 - 2a^2 + c^2 + 2b^3=a^2+7b^3+c^2{{/formula}} |
| 2 | 1.b) {{formula}}2xy^2 + 8x^2 + y^2x - 2x^2 + xy^2 + 2y^2x=3xy^2+6x^2+3y^2x{{/formula}} | ||
| 3 | 1.c) | ||
| 4 | |||
| 5 | {{formula}} | ||
| 6 | \begin{align} | ||
| 7 | &2(4x)^2 + 2 - 6x^2 - (3x)^2 - 6x - 1 \\ | ||
| 8 | &=2(16x^2)+2-6x^2-9x^2-6x-1\\ | ||
| 9 | &=32x^2+1-15x^2-6x\\ | ||
| 10 | &=17x^2-6x+1 | ||
| 11 | \end{align} | ||
| 12 | {{/formula}} | ||
| 13 | |||
| 14 | Wende die Potenzgesetze an: | ||
| 15 | 2.a) {{formula}}a^2 \cdot a^4 + b \cdot b^5=a^6+b^6{{/formula}} | ||
| 16 | |||
| 17 | 2.b) {{formula}}-10a^2 + 2a(a+2)=-10a^2+2a^2+4a=-8a^2+4a{{/formula}} | ||
| 18 | |||
| 19 | 2.c) {{formula}}y^3 \cdot (-x)^3=(y\cdot(-x))^3=(-xy)^3{{/formula}} | ||
| 20 | |||
| 21 | 2.d) {{formula}}\left(\frac{x}{3}\right)^4 \cdot 3^4=\left(\frac{x}{3}\cdot 3\right)^4=x^4{{/formula}} | ||
| 22 | |||
| 23 | 2.e) {{formula}}\frac{b^{n+2}}{b^n}=b^{n+2-n}=b^2{{/formula}} | ||
| 24 | |||
| 25 | 2.f) {{formula}}\frac{(2x)^5}{(2x)^{a+5}}=(2x)^{5-a-5}=(2x)^{-a}{{/formula}} | ||
| 26 | |||
| 27 | 2.g) {{formula}}\frac{2^3}{\left(\frac{1}{2}\right)^3}=\left(\frac{2}{\frac{1}{2}}\right)=4^3=64{{/formula}} | ||
| 28 | |||
| 29 | 2.h) {{formula}}\frac{(-2x)^4}{(-y)^4}=\left(\frac{-2x}{-y}\right)^4=\frac{16x^4}{y^4}{{/formula}} | ||
| 30 | |||
| 31 | 2.i) {{formula}}(-2y)^3=(-2)^3\cdot y^3=-8y^3{{/formula}} | ||
| 32 | |||
| 33 | 2.j) {{formula}}(5a^3b^2)^3=5^3\cdot a^{3\cdot3}\cdot b^{2\cdot 3}=125a^9b^6{{/formula}} |