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1/ a/ \(\sqrt{0,9.0,16.0,4}=\sqrt{\frac{9.16.4}{10000}}=\sqrt{\frac{\left(3.4.2\right)^2}{10^4}}=\frac{24}{1010}=\frac{6}{25}\)
b/ \(\sqrt{0,0016}=\sqrt{\frac{16}{100}}=\frac{4}{10}=\frac{2}{5}\)
c/ \(\frac{\sqrt{72}}{\sqrt{2}}=\frac{\sqrt{2}.\sqrt{36}}{\sqrt{2}}=\sqrt{36}=6\)
d/ \(\frac{\sqrt{2}}{\sqrt{288}}=\frac{\sqrt{2}}{\sqrt{2}.\sqrt{144}}=\frac{1}{\sqrt{144}}=\frac{1}{12}\)
2.
a/ \(\frac{2}{a}.\sqrt{\frac{16a^2}{9}}=\frac{2}{a}.\frac{4\left|a\right|}{3}=-\frac{8a}{3a}=-\frac{8}{3}\) (Vì a<0)
b/ \(\frac{3}{a-1}.\sqrt{\frac{4a^2-8a+4}{25}}=\frac{3}{a-1}.\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3.2\left|a-1\right|}{5.\left(a-1\right)}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{\sqrt{243a}}{\sqrt{3a}}=\frac{9\sqrt{3a}}{\sqrt{3a}}=9\)
d/ \(\frac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}=\frac{3.3\sqrt{2}.\left|a\right|.\left|b\right|^2}{\sqrt{2}.\left|a\right|.\left|b\right|}=9\left|b\right|\)
a/ \(\sqrt{8\left(\sqrt{2}-\sqrt{3}\right)^2}=2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{6}-4\)
b/ \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab.\sqrt{\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2.\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2+1}\)
c/ \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a}{b^3}\left(1+\frac{1}{b}\right)}=\frac{1}{b}.\sqrt{\frac{a}{b}\left(1+\frac{1}{b}\right)}\)
d/ \(\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)
Cố gắng hơn nữa ah. Thế vô là thấy nó sai liền nên m không giải nữa.
Thay \(\hept{\begin{cases}a=2\\b=2\end{cases}}\) thì ta có:
\(\left(\sqrt[3]{2^4}+2^2.\sqrt[3]{2^2}+2^4\right).\frac{\left(\sqrt[3]{2^8}-2^6+2^4.\sqrt[3]{2^2}-2^2.2^2\right)}{2^2.2^2+2^2-2^8.2^2-2^4}=2^2.2^2\)
\(\Leftrightarrow1,477=16\left(sai\right)\)
Vậy đề bài cho tào lao.