Question

\(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)^2}\)

a. tính gt P khi x = \(\frac{\sqrt{7-4\sqrt{3}}}{2}\)

b. tìm x sao cho P = 1/2

giúp mình vs ạ, mình cần gấp, cảm ơn trc ạ

Answer 1

Lời giải:

ĐK: $x\geq 0$

a)

Khi \(x=\frac{\sqrt{7-4\sqrt{3}}}{2}=\frac{\sqrt{4+3-2\sqrt{4.3}}}{2}=\frac{\sqrt{(2-\sqrt{3})^2}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{(\sqrt{3}-1)^2}{2^2}\)

\(\Rightarrow \sqrt{x}=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow \left\{\begin{matrix} 4\sqrt{x}=2(\sqrt{3}-1)\\ (\sqrt{x}+1)^2=\frac{4+2\sqrt{3}}{4}\end{matrix}\right.\) \(\Rightarrow P=-20+12\sqrt{3}\)

b)

\(P=\frac{4\sqrt{x}}{(\sqrt{x}+1)^2}=\frac{1}{2}\)\(\Leftrightarrow 8\sqrt{x}=x+1+2\sqrt{x}\)

\(\Leftrightarrow x-6\sqrt{x}+1=0\)

\(\Leftrightarrow (\sqrt{x}-3)^2=8\Rightarrow \sqrt{x}-3=\pm 2\sqrt{2}\)

\(\Rightarrow \sqrt{x}=3-2\sqrt{2}\Rightarrow x=17\pm 12\sqrt{2}\)

(đều thỏa mãn)

Answer 2

Lời giải:

ĐK: $x\geq 0$

a)

Khi \(x=\frac{\sqrt{7-4\sqrt{3}}}{2}=\frac{\sqrt{4+3-2\sqrt{4.3}}}{2}=\frac{\sqrt{(2-\sqrt{3})^2}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{(\sqrt{3}-1)^2}{2^2}\)

\(\Rightarrow \sqrt{x}=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow \left\{\begin{matrix} 4\sqrt{x}=2(\sqrt{3}-1)\\ (\sqrt{x}+1)^2=\frac{4+2\sqrt{3}}{4}\end{matrix}\right.\) \(\Rightarrow P=-20+12\sqrt{3}\)

b)

\(P=\frac{4\sqrt{x}}{(\sqrt{x}+1)^2}=\frac{1}{2}\)\(\Leftrightarrow 8\sqrt{x}=x+1+2\sqrt{x}\)

\(\Leftrightarrow x-6\sqrt{x}+1=0\)

\(\Leftrightarrow (\sqrt{x}-3)^2=8\Rightarrow \sqrt{x}-3=\pm 2\sqrt{2}\)

\(\Rightarrow \sqrt{x}=3-2\sqrt{2}\Rightarrow x=17\pm 12\sqrt{2}\)

(đều thỏa mãn)