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Question

Tính :

1) $\frac{1}{6}^{\log _62-\frac{1}{2}\log _\sqrt{6}5 }$

2) $\frac{1}{16}^{-0,75}+(0,25)^{-\frac{5}{2}}$ = 40

3) $\left ( \frac{b^2}{a} \right )^{-2}.\left ( \frac{a^2}{b} \right )^{-3}$

4) $\frac{(\sqrt[3]{4a} )^2}{\sqrt[6]{4a} }$

5) $\frac{a^\frac{4}{3}\left ( a^{-\frac{1}{3}}+a^\frac{2}{3}\right )}{a^\frac{1}{4}\left ( a^\frac{3}{4} +a^{-\frac{1}{4}}\right )}$

khangnguyenthanh · Answer 1 · 7/27/2013 4:23:50 PM

3. $\left(\dfrac{b^2}{a}\right)^{-2}.\left(\dfrac{a^2}{b}\right)^{-3}$

$=\dfrac{b^{-4}}{a^{-2}}.\dfrac{a^{-6}}{b^{-3}}$

$=\dfrac{b^{-1}}{a^{-4}}=\dfrac{1}{a^4b}$

4. $\dfrac{(\sqrt[3]{4a})^2}{\sqrt[6]{4a}}$

$=\dfrac{(4a)^{\frac{2}{3}}}{(4a)^{\frac{1}{6}}}$

$=(4a)^{\frac{1}{2}}=2\sqrt a$

5. $\dfrac{a^{\frac{4}{3}}(a^{\frac{-1}{3}}+a^{\frac{2}{3}})}{a^{\frac{1}{4}}(a^{\frac{3}{4}}+a^{\frac{-1}{4}})}$

$=\dfrac{a+a^2}{a+1}=a$

Trả lời 27-07-13 04:23 PM

khangnguyenthanh
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khangnguyenthanh · Answer 2 · 7/27/2013 4:17:25 PM

2. $\left(\dfrac{1}{16}\right)^{-0.75}+(0.25)^{\frac{-5}{2}}$

$=\left(\dfrac{1}{2}\right)^{-3}+\left(\dfrac{1}{2}\right)^{-5}$

$=8+32=40$

Trả lời 27-07-13 04:17 PM

khangnguyenthanh
72K 2 480 42

332K 4M

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khangnguyenthanh · Answer 3 · 7/27/2013 4:12:43 PM

1. $\dfrac{1}{6}^{\log_62-\frac{1}{2}\log_{\sqrt6}5}$

$=\dfrac{1}{6}^{\log_62-\log_65}$

$=\dfrac{1}{6}^{\log_6\frac{2}{5}}$

$=\dfrac{-2}{5}$

Trả lời 27-07-13 04:12 PM

khangnguyenthanh
72K 2 480 42

332K 4M

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