Question

Rút gọn:

N = \(\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}\)

P = \(\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)(\(x\ge0\),\(x\ne4\))

Answer 1

+) ta có : \(N=\dfrac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\dfrac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}\left(\sqrt{30}-\sqrt{2}\right)}=\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}\)

\(=\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\dfrac{1}{2}\)

+) ta có : \(P=\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)

\(\Leftrightarrow P=\left(\dfrac{\left(2-\sqrt{x}\right)\left(4+2\sqrt{x}+x\right)}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\) \(\Leftrightarrow P=\left(4+2\sqrt{x}+x+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\) \(\Leftrightarrow P=\left(2+\sqrt{x}\right)^2\dfrac{\left(2-\sqrt{x}\right)^2}{\left(2+\sqrt{x}\right)^2}=\left(2-\sqrt{x}\right)^2\)

Answer 2

<=>N=\(\dfrac{\sqrt{16-2\sqrt{15}}}{\sqrt{60}-2}\)

<=>N=\(\dfrac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\sqrt{15}-2}\)

<=>N=\(\dfrac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}\)

<=>N=\(\dfrac{1}{2}\)

P=\(\left(\dfrac{8-x\sqrt{x}}{2-\sqrt{x}}+2\sqrt{x}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)

P=\(\left(\dfrac{8-x\sqrt{x}+4\sqrt{x}-2x}{2-\sqrt{x}}\right)\left(\dfrac{2-\sqrt{x}}{2+\sqrt{x}}\right)^2\)

P=\(\dfrac{8+3\sqrt{x}+x}{2-\sqrt{x}}.\dfrac{\left(2-\sqrt{x}\right)^2}{\left(2+\sqrt{x}\right)^2}\)

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