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

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

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

= 5+(-7) - 2.4 - \(\dfrac{1}{3}\).6

= - 12

a. ĐKXĐ: x < 2

a, \(\left\{{}\begin{matrix}2x-1\ne0\\\frac{x^2}{2x-1}\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{1}{2}\\2x-1>0\end{matrix}\right.\Leftrightarrow x>\frac{1}{2}\)

b, \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}-\sqrt[3]{-216}.\sqrt[3]{\frac{1}{27}}=\frac{\sqrt[3]{5^3.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-6\right)^3}.\sqrt[3]{\left(\frac{1}{3}\right)^3}\)

\(=\frac{5\sqrt[3]{5}}{\sqrt[3]{5}}+6.\frac{1}{3}=5+2=7\)

a, = \(3\sqrt{2}+3\sqrt{3}+9\sqrt{2}-5\sqrt{3}\)

= \(12\sqrt{2}-2\sqrt{3}\)

b, = 5 - 7 - 8 + 2

= - 8

Bài 1 :

a, ĐKXĐ : \(\dfrac{1}{2-x}\ge0\)

Mà 1 > 0

\(\Rightarrow2-x>0\)

\(\Rightarrow x< 2\)

Vậy ...

b, Ta có : \(\sqrt[3]{125}.\sqrt[3]{216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}\)

\(=5.6-\dfrac{8.1}{2}=26\)

1a) Để căn thức bậc 2 có nghĩa thì \(\dfrac{1}{2-x}\ge0\Rightarrow2-x>0\Rightarrow x< 2\)

b) \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}=\sqrt[3]{5^3}.\sqrt[3]{\left(-6\right)^3}-\sqrt[3]{8^3}.\sqrt[3]{\left(\dfrac{1}{2}\right)^3}\)

\(=5.\left(-6\right)-8.\dfrac{1}{2}=-34\)

\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{b}\right)^2}-\dfrac{\sqrt{a}}{\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}-\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(=-\dfrac{\sqrt{b}}{\sqrt{b}}=-1< 0\)

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=3\sqrt{3}\)

vậy \(A=3\sqrt{3}\)

\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)  \(ĐKXĐ:x>0;x\ne1\)

\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)

\(B=1-x\)

vậy \(B=1-x\)

\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)

\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)

\(C=4+5+6\)

\(C=15\)

vậy \(C=15\)

Cho mk giải câu a:

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=\left(18-10-5\right)\sqrt{3}\)

\(A=3\sqrt{3}\)