Question

[Park Chanyeol]

Rút gọn

\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\) -( \(\sqrt{x}-\sqrt{y}\))²

\(\sqrt{\frac{x-2\sqrt{x}}{x+2\sqrt{x}+1}}\) (x >_ 0)
\(\frac{x-1}{\sqrt{y}-1}\) . \(\sqrt{\frac{\left(2\sqrt{y}+1\right)^2}{\left(x-1\right)}}\) với x # 1, y# 1,y>0
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}\) : \(\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}\) rồi tính giá trị với a= 7,25 b= 3,25
4x - \(\sqrt{8}\) + \(\sqrt{\frac{x^3+2x^2}{\sqrt{x+2}}}\) với x =- \(\sqrt{2}\)

Answer 1

[Nguyễn Ngọc Lộc]

a, ĐKXĐ : \(\left[{}\begin{matrix}x\ge0\\ y>0\end{matrix}\right.\) hoặc \(\left[{}\begin{matrix}x>0\\y\ge0\end{matrix}\right.\)

Ta có :\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

= \(\frac{\sqrt{x^2}\sqrt{x}+\sqrt{y^2}\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

= \(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\left(x-2\sqrt{xy}+y\right)\)

= \(\left(x-\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\)

= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

= \(\sqrt{xy}\)

Answer 2

[Chén Thánh]

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

Thay a=7,25 và b= 3,25 vào (*) ta có:

\(\sqrt{\frac{7,25^2-1}{3,25^2-1}}\) \(=\frac{5\sqrt{33}}{4}:\frac{3\sqrt{17}}{4}=\frac{5\sqrt{33}}{3\sqrt{17}}=\frac{5\sqrt{561}}{51}\)