a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)

\(=\dfrac{2}{3}\)

a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)

\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)

=2

b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)

\(=\dfrac{4x-1}{x^2}\)

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

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

\(=-a^2\)

\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)

\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)

\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)

\(B=\left(\dfrac{1}{\sqrt[]{a}-1}-\dfrac{1}{\sqrt[]{a}}\right):\left(\dfrac{\sqrt[]{a}+1}{\sqrt[]{a}-2}-\dfrac{\sqrt[]{a}+2}{\sqrt[]{a}-1}\right)\left(1\right)\)

a) B xác định khi và chỉ khi :

\(\left\{{}\begin{matrix}a\ge0\\\sqrt[]{a}\ne0\\\sqrt[]{a}-1\ne0\\\sqrt[]{a}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)

b) \(\left(1\right)\Leftrightarrow B=\left(\dfrac{\sqrt[]{a}-\left(\sqrt[]{a}-1\right)}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{\left(\sqrt[]{a}+1\right)\left(\sqrt[]{a}-1\right)-\left(\sqrt[]{a}+2\right)\left(\sqrt[]{a}-2\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)

\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{a-1-\left(a-4\right)}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)

\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right):\left(\dfrac{3}{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}\right)\)

\(\Leftrightarrow B=\left(\dfrac{1}{\sqrt[]{a}\left(\sqrt[]{a}-1\right)}\right).\left(\dfrac{\left(\sqrt[]{a}-1\right)\left(\sqrt[]{a}-2\right)}{3}\right)\)

\(\Leftrightarrow B=\dfrac{\sqrt[]{a}-2}{3\sqrt[]{a}}\)

\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)

\(=\pm3\sqrt{2}x^2\)

\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)

\(=\pm\left|a\right|\)

a: ĐKXĐ: x<>1/2

Sửa đề: \(A=\frac{3}{2\left(2x-1\right)}\cdot\sqrt{8x^4\left(4x^2-4x+1\right)}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot\sqrt8\cdot\sqrt{x^4}\cdot\sqrt{\left(2x-1\right)^2}\)

\(=\frac{3}{2\left(2x-1\right)}\cdot2\sqrt2\cdot x^2\cdot\left|2x-1\right|=\frac{6\sqrt2\cdot x^2}{2\left(2x-1\right)}\cdot\left|2x-1\right|\)

=\(\pm3\sqrt2\cdot x^2\)

b: ĐKXĐ: b<>0

\(B=\frac{a-b}{b^2}\cdot\sqrt{\frac{a^2b^4}{a^2-2ab+b^2}}\)

\(=\frac{a-b}{b^2}\cdot\sqrt{a^2}\cdot\frac{\sqrt{b^4}}{\sqrt{\left(a-b\right)^2}}\)

\(=\frac{a-b}{b^2}\cdot\left|a\right|\cdot\frac{b^2}{\left|a-b\right|}=\left|a\right|\cdot\frac{a-b}{\left|a-b\right|}=\pm\left|a\right|\)

a: Sửa đề: \(B=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

ĐKXĐ: a>0; a<>1

b: Ta có: \(B=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)}\)

\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+2\sqrt{a}+1-a+2\sqrt{a}-1}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\cdot4\sqrt{a}}=\frac{\sqrt{a}+1}{4a}\)