4 replies to this topic

[trucvan]

Cho $5<x\leq10$ và $\sqrt{x}+\sqrt{10-x}$=k. Tính giá trị biểu thức B=$\frac{\sqrt{5-\sqrt{10x-x^{2}}}}{x-5}$

Edited by trucvan, 19-07-2017 - 12:46.

[trieutuyennham]

Do $\sqrt{x}+\sqrt{10-x}=k$ nên $\sqrt{10x-x^{2}}=\frac{k^{2}}{2}-5$

Ta có 

$B^{2}=\frac{10-\frac{k^{2}}{2}}{25-(\frac{k^{2}}{2}-5)^{2}}=\frac{2}{k^{2}}$

$\Rightarrow B=\frac{\sqrt{2}}{k}$

[CunHenry01]

$/frac{a}{b+c}$

[CunHenry01]

$\frac{a}{b+c}$

[Twogvecman]

$$ Test $$

$$ a_2 ^ 2 $$

Edited by Twogvecman, 17-08-2017 - 16:07.