a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)

b) Với \(a>0;a\ne1;a\ne4\), ta có:

\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).

a, ĐKXĐ: 

\(\left\{{}\begin{matrix}\left|a\right|>1^2\\\left|a\right|>0\\\left|a\right|>2^2\end{matrix}\right.\Leftrightarrow a>4\)

b,

 \(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ B=\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left[\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)\right]}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ B=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}\\ B=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

\(c,B\le\dfrac{1}{3}\\ \Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\\ \Leftrightarrow3\left(\sqrt{a}-2\right)\le3\sqrt{a}\\ \Leftrightarrow\sqrt{a}-2\le\sqrt{a}\\ \Leftrightarrow\sqrt{a}-\sqrt{a}\le2\\ \Leftrightarrow0\le2\left(luôn.đúng\right)\)

Vậy: Với a>4 thì \(B\le\dfrac{1}{3}\)

a)ĐKXĐ:\(\begin{cases}x\ge0\\2\sqrt{x}-2\ne0\\1-x\ne0\\\end{cases}\)

`<=>` \(\begin{cases}x\ge0\\x\ne1\\\end{cases}\)

`B=1/(2sqrtx-2)-1/(2sqrtx+2)+sqrtx/(1-x)`

`=1/(2(sqrtx-1))-1/(2(sqrtx+1))-sqrtx/(x-1)`

`=(sqrtx+1-(sqrtx-1)-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`

`=(2-2sqrtx)/(2(sqrtx-1)(sqrtx+1))`

`=(2(1-sqrtx))/(2(sqrtx-1)(sqrtx+1))`

`=-1/(sqrtx+1)`

`b)x=3`

`=>B=(-1)/(sqrt3+1)`

`=(-(sqrt3-1))/(3-1)`

`=(1-sqrt3)/2`

`c)|A|=1/2`

`<=>|(-1)/(sqrtx+1)|=1/2`

`<=>|1/(sqrtx+1)|=1/2`

`<=>1/(sqrtx+1)=1/2` do `1>0,sqrtx+1>=1>0`

`<=>sqrtx+1=2`

`<=>sqrtx=1`

`<=>x=1` loại vì `x ne 1`.

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)

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

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

\(=\dfrac{-1}{\sqrt{x}+1}\)

b) Thay x=3 vào B, ta được:

\(B=\dfrac{-1}{\sqrt{3}+1}=\dfrac{-\sqrt{3}+1}{2}\)

c) Ta có: \(\left|A\right|=\dfrac{1}{2}\)

nên \(\left[{}\begin{matrix}A=\dfrac{1}{2}\\A=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{-1}{\sqrt{x}+1}=\dfrac{1}{2}\\\dfrac{-1}{\sqrt{x}+1}=\dfrac{-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=-2\\\sqrt{x}+1=2\end{matrix}\right.\Leftrightarrow x=1\)(loại)

Lời giải:

a. ĐKXĐ: $x>1$

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

b.

\(B=\frac{a^2+b^2}{2ab}+\sqrt{\frac{a^2+2ab+b^2}{2ab}.\frac{a^2-2ab+b^2}{2ab}}\)

\(=\frac{a^2+b^2}{2ab}+\sqrt{\frac{(a+b)^2(a-b)^2}{(2ab)^2}}=\frac{a^2+b^2}{2ab}+\frac{|a-b||a+b|}{|2ab|}=\frac{a^2+b^2}{2ab}+\frac{a^2-b^2}{2ab}=\frac{a}{b}\)

c.

$B\leq 1\Leftrightarrow (x-1)+\sqrt{x^2-1}\leq 0$

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-1}+\sqrt{x+1})\leq 0$

$\Leftrightarrow \sqrt{x-1}\leq 0$

Mà $\sqrt{x-1}>0$ với mọi $x<1$ nên điều này vô lý)

Vậy không tồn tại $x$ thỏa đkđb

d.

$B=2\Leftrightarrow x+\sqrt{x^2-1}=2$

$\Leftrightarrow \sqrt{x^2-1}=2-x$

\(\Rightarrow \left\{\begin{matrix} 2-x\geq 0\\ x^2-1=(2-x)^2=x^2-4x+4\end{matrix}\right.\)

\(\Rightarrow x=\frac{5}{4}\)

Thử lại thấy thỏa mãn

Vậy......

ĐKXĐ : \(\left\{{}\begin{matrix}a^2-b^2>0\\a-\sqrt{a^2-b^2}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2>b^2\\a^2-b^2\ne a^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2>b^2\\b^2\ne0\end{matrix}\right.\)

Lời giải:
ĐKXĐ: \(\left\{\begin{matrix} a^2-b^2>0\\ a-\sqrt{a^2-b^2}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a^2>b^2\\ a\neq \sqrt{a^2-b^2}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a^2> b^2\\ a\neq \sqrt{a^2-b^2}\end{matrix}\right.\)

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)