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a. \(\sqrt{\frac{y}{5x^3}}=\sqrt{\frac{5xy}{25x^4}}=\frac{\sqrt{5xy}}{25x^2}\)
b\(\sqrt{\frac{5}{x\left(1-\sqrt{2}\right)}}=\sqrt{\frac{5\times x\left(1+\sqrt{2}\right)}{x^2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}}=\sqrt{\frac{-5\times x\left(1+\sqrt{2}\right)}{x^2}}=-\frac{\sqrt{-5\times x\left(1+\sqrt{2}\right)}}{x}\)
c.\(\sqrt{\frac{x-1}{2\left(\sqrt{x}-1\right)}}=\sqrt{\frac{\sqrt{x}+1}{2}}=\frac{\sqrt{2\sqrt{x}+2}}{2}\)
d.\(a\sqrt{\frac{4}{a}}=\sqrt{\frac{4a^2}{a}}=\sqrt{4a}=2\sqrt{a}\)
e.\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\left(\text{ do a< 0}\right)\)\(2\sqrt{\frac{1}{-a}}=2\sqrt{\frac{-a}{a^2}}=-\frac{2}{a}\sqrt{-a}\)( do a <0)
f.\(\sqrt{\frac{2}{x-1}-\frac{1}{\left(x-1\right)^2}}=\sqrt{\frac{2\left(x-1\right)-1}{\left(x-1\right)^2}}=\frac{\sqrt{2x-3}}{\left|x-1\right|}\)
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)
Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)
= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
= \(\sqrt{xy}\)
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) \(=\sqrt{\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}+1\right)\left(\sqrt{b}-1\right)}}\)\(=\sqrt{\frac{a^2-1}{b^2-1}}\) (*)
Thay a=7,25 và b= 3,25 vào (*) ta có:
\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
a)\(=-\sqrt{\left(\frac{a}{b}\right)^2\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a^2}{b^2}\cdot\frac{b}{a}}\)
\(=-\sqrt{\frac{a}{b}}\)
b) \(=\sqrt{\left(\frac{1}{2x-1}\right)^2\cdot5\left(4x^2-4x+1\right)}\)
\(=\sqrt{\frac{5}{\left(2x-1\right)^2}\cdot\left(2x-1\right)^2}\)
\(=\sqrt{5}\)