Bài 2:

ĐKXĐ: \(x\geq 0; x\neq 1\)

a)

\(B=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3}{x-1}-\frac{6\sqrt{x}-4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Khi \((x^2+1)(2x-8)=0\Rightarrow \left[\begin{matrix} x^2+1=0\\ 2x-8=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x^2=-1(\text{vô lý})\\ x=4(\text{thỏa mãn})\end{matrix}\right.\)

Với $x=4$:

\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)

Bài 1:

ĐKXĐ: \(x\geq 0; x\neq 4\)

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

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

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

b)

Khi \(|x|=25\Rightarrow \left[\begin{matrix} x=25\\ x=-25\end{matrix}\right.\). Mà $x\geq 0$ nên $x=25$

\(P=\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{3}\)

Bài 2:

ĐKXĐ: \(x\geq 0; x\neq 1\)

a)

\(B=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3}{x-1}-\frac{6\sqrt{x}-4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Khi \((x^2+1)(2x-8)=0\Rightarrow \left[\begin{matrix} x^2+1=0\\ 2x-8=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x^2=-1(\text{vô lý})\\ x=4(\text{thỏa mãn})\end{matrix}\right.\)

Với $x=4$:

\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)

a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)

\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)

\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)

\(=-\left(2-5\right)\)

\(=-\left(-3\right)\)

\(=3\)

b) Ta có:

\(x^2-x\sqrt{3}+1\) 

\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)

Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra:

\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)

Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)

a)

\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)