a) \(\sqrt{\dfrac{x}{y^3}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy}{y^4}+\dfrac{2x}{y^4}}=\sqrt{\dfrac{xy+2x}{y^4}}=\dfrac{\sqrt{xy+2x}}{\sqrt{y^4}}=\dfrac{\sqrt{xy+2x}}{\left|y^2\right|}=\dfrac{\sqrt{xy+2x}}{y^2}\)(vì y²\(\ge0\))

b) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\sqrt{x}-\sqrt{x}.\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

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

Nếu a-b>0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{a-b}=\left|ab\right|\)

Nếu a-b<0 thì \(\left(a-b\right)\dfrac{\left|ab\right|}{\left|a-b\right|}=\left(a-b\right)\dfrac{\left|ab\right|}{-\left(a-b\right)}=-\left|ab\right|\)

d) \(\dfrac{a-3\sqrt{a}+3}{a\sqrt{a}+3\sqrt{3}}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}\right)^3+\left(\sqrt{3}\right)^3}=\dfrac{a-3\sqrt{a}+3}{\left(\sqrt{a}+\sqrt{3}\right)\left(a-3\sqrt{a}+3\right)}=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)

Nếu trục căn thức ở mẫu thì \(\dfrac{1}{\sqrt{a}+\sqrt{3}}=\dfrac{\sqrt{a}-\sqrt{3}}{\left(\sqrt{a}+\sqrt{3}\right)\left(\sqrt{a}-\sqrt{3}\right)}=\dfrac{\sqrt{a}-\sqrt{3}}{a-3}\)

đây là b) nhé

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

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

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

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

Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???

b.

\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)

\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)

\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)

Nguyễn Hoàng trung: Chả qua nếu $a=6-3\sqrt{3}; b=2+\sqrt{3}$ thì kết quả sẽ đẹp hơn. Còn như đề thì vẫn rút gọn được.

\(A=\frac{a-\sqrt{ab}+b}{(\sqrt{a}-\sqrt{b})^2}=\frac{a-\sqrt{ab}+b}{a-2\sqrt{ab}+b}\)

\(2a^2=12-6\sqrt{3}=(3-\sqrt{3})^2\Rightarrow a=\frac{3-\sqrt{3}}{\sqrt{2}}\) (do $a\geq 0$)

\(2b^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow b=\frac{\sqrt{3}+1}{\sqrt{2}}\) (do $b\geq 0$)

\(\Rightarrow a+b=2\sqrt{2}; ab=\frac{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}{2}=\sqrt{3}\)

Do đó: $A=\frac{2\sqrt{2}-\sqrt[4]{3}}{2\sqrt{2}-2\sqrt[4]{3}}$

@Nhã Doanh

a) \(\dfrac{x-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

a: \(=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}\)

b: \(=\left(a-b\right)\cdot\dfrac{\left|ab\right|}{\left|a-b\right|}=\pm\left|ab\right|\)

c: \(=\dfrac{1}{\sqrt{a}+\sqrt{3}}\)

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

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

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