1.a) Để căn thức có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2}{2x-1}\ge0\\2x-1\ne0\end{matrix}\right.\)

\(\Leftrightarrow2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)

Vậy...

b, \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\dfrac{1}{27}}=\sqrt[3]{\dfrac{625}{5}}-\sqrt[3]{-\dfrac{216}{27}}=\sqrt[3]{125}-\sqrt[3]{-8}=5-\left(-2\right)=7\)

a) Để căn thức có nghĩa thì 2x-1>0

\(\Leftrightarrow2x>1\)

hay \(x>\dfrac{1}{2}\)

b) Ta có: \(\dfrac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}\cdot\sqrt[3]{\dfrac{1}{27}}\)

\(=5-\left(-6\right)\cdot\dfrac{1}{3}\)

\(=5+6\cdot\dfrac{1}{3}=5+2=7\)

a, \(=>3-\sqrt{2}+\sqrt{50}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}\)

b, \(=>\dfrac{\sqrt[3]{125.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-4\right).2}=\sqrt[3]{125}-\sqrt[3]{\left(-2\right)^3}\)

\(=5-\left(-2\right)=7\)

c, \(=>\sqrt{6}.\sqrt{\dfrac{6}{2}}-\sqrt{2}-3\sqrt{4.2}=\sqrt{6}.\sqrt{3}-\sqrt{2}-6\sqrt{2}\)

\(=\sqrt{18}-7\sqrt{2}=3\sqrt{2}-7\sqrt{2}=-4\sqrt{2}\)

d, \(=>\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-\dfrac{2}{\sqrt{3}-1}\)

\(=\dfrac{3-\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{1-\sqrt{3}}{\sqrt{3}-1}=-1\)

ĐK: \(n\le\dfrac{625}{4}\le156\) (vì \(n\in Z\) )

Đặt \(a=\sqrt{\dfrac{25}{2}+\sqrt{\dfrac{625}{4}-n}}+\sqrt{\dfrac{25}{2}-\sqrt{\dfrac{625}{4}-n}}\) \(\left(a\ge0,a\in Z\right)\)

\(\Rightarrow a^2=25+2\sqrt{\dfrac{625}{4}-\dfrac{625}{4}+n}\)

\(\Rightarrow a^2=25+2\sqrt{n}\) (1)

Để \(a\in Z\Rightarrow a^2\in Z\Rightarrow\sqrt{n}\in Z^+\)

Vì \(2\sqrt{n}⋮2\) mà 25 không chia hết cho 2

\(\Rightarrow a^2\) không chia hết cho 2

\(\Rightarrow\) a không chia hết cho 2

Đặt \(a=2k+1\left(k>0,k\in Z\right)\)

\(\left(1\right)\Rightarrow\left(2k+1\right)^2=25+2\sqrt{n}\)

\(\Rightarrow2\sqrt{n}=4k^2+4k-24\)

\(\Rightarrow\sqrt{n}=2k^2+2k-12\)

Vì \(\sqrt{n}\ge0\Rightarrow2k^2+2k-12\ge0\)

\(\Rightarrow\left(k+3\right)\left(k-2\right)\ge0\)

Vì \(k>0\Rightarrow k\ge2\) (2)

Mặt khác: \(n\le156\Rightarrow\sqrt{n}\le\sqrt{156}\) mà \(\sqrt{n}\in Z\)

\(\Rightarrow\sqrt{n}\le12\Rightarrow2k^2+2k-12\le12\)

\(\Rightarrow\left(k-3\right)\left(k+4\right)\le0\)

Vì \(k>0\Rightarrow0< k\le3\) (3)

Từ (2) và (3)\(\Rightarrow\left[{}\begin{matrix}k=2\\k=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}n=0\\n=144\end{matrix}\right.\) (t/m)

Vậy n=0, n=144

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\(\sqrt{484}-\dfrac{1}{\sqrt{5}}< \sqrt{529}-\dfrac{1}{19}< \sqrt{576}-\dfrac{1}{\sqrt{7}}< \sqrt{625}-\dfrac{1}{\sqrt{8}}\)

\(\left(\dfrac{1}{\sqrt{625}}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{\sqrt{25}}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\left(-\dfrac{29}{25}\right)\)
\(=\dfrac{31}{25}.\left(-\dfrac{25}{29}\right)\)
\(=-\dfrac{31}{29}\)