Question

[Hà Ngân Hà]

a. Cho biểu thức \(B=\dfrac{2}{\sqrt{x}-2}\) với x≥0, x≠4. Tìm x biết rằng \(B=\dfrac{1}{\sqrt{3}+2}+\dfrac{1}{\sqrt{3}-2}\)

b. Cho biểu thức \(A=\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\). Rút gọn A và tính \(P=\dfrac{B}{A}\)

c. Tìm x thỏa mãn: \(P.\left(\sqrt{x}+1\right)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)

Answer 1

[Akai Haruma]

Lời giải:

a)

Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)

Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)

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

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

\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)

b)

ĐK: \(x\geq 0; x\neq 4\)

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

\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)

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

Answer 2

[Akai Haruma]

c) Thêm ĐK: \(x\geq 1\)

Từ biểu thức P vừa tìm được:

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

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

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

\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)

\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)

Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)

\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :

\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)

Vậy..........