Pierwiastki, pot臋gi, logarytmy, zadanie nr 2687
ostatnie wiadomo艣ci | regulamin | latex
| Autor | Zadanie / Rozwi膮zanie |
angela post贸w: 131 |  2013-04-02 22:24:301) zapisz w postaci jednej pot臋gi o wyk艂adniku wymiernym $5\sqrt{5\sqrt{5}}$ $4\sqrt{2\cdot\sqrt[3]{2}}$ 2) oblicz a)$[(4-12^\frac{1}{2})^\frac{1}{2}+(4+12^\frac{1}{2})^\frac{1}{2}]^2$ b)$\frac{log\sqrt{128}+log 32^\frac{1}{3}}{log(2\sqrt{2})}$ c)wiedz膮c ze log12=a i log2=b oblicz log90 |
tumor post贸w: 8070 | 2013-04-02 22:29:19 $ 5\sqrt{5\sqrt{5}}=5(5*5^\frac{1}{2})^\frac{1}{2}=5*5^\frac{3}{4}=5^\frac{7}{4}$ |
tumor post贸w: 8070 | 2013-04-02 22:29:31 $4\sqrt{2*\sqrt[3]{2}}=2^2*2^\frac{1}{2}*2^\frac{1}{6}=2^\frac{8}{3}$ |
tumor post贸w: 8070 | 2013-04-02 22:34:08 2) a) = $(4-12^\frac{1}{2})+2 (4-12^\frac{1}{2})^\frac{1}{2}(4+12^\frac{1}{2})^\frac{1}{2}+(4+12^\frac{1}{2})=4+4+2(16-12)^\frac{1}{2}=12$ |
agus post贸w: 2387 | 2013-04-02 22:55:01 2b) =$\frac{log (2^{7})^{\frac{1}{2}} \cdot (2^{5})^{\frac{1}{3}}}{log 2^{\frac{3}{2}}}$= $\frac{log 2^{\frac{21}{6}} \cdot 2^{\frac{10}{6}}}{log 2^{\frac{9}{6}}}$= =$\frac{\frac{31}{6}log2}{\frac{9}{6}log2}=\frac{31}{9}$ |
agus post贸w: 2387 | 2013-04-02 23:03:22 log90=log(10$\cdot$9)=log10+log9=1+log$3^{2}$=1+2log3 a=log12=log($2^{2}\cdot3$)=log$2^{2}$+log3=2log2+log3=2b+log3 a=2b+log3 log3=a-2b 1+2log3=1+2(a-2b)=2a-4b+1 |
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