\(A=4x-\sqrt{4x^2-12x+9}\)

\(=4x-2x+3\)

\(=2x+3\)

\(A=15\Rightarrow2x+3=15\)

\(2x=12\)

\(x=6\)

a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)

b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))

`đk:x-\sqrt{x^2-4x+4}>=0`

`<=>x>=\sqrt{x^2-4x+4}`

`<=>x^2>=x^2-4x+4(x>=0)`

`<=>4x-4>=0`

`<=>4x>=4<=>x>=1`

`b)A=sqrt{x-sqrt{(x-2)^2}}`

`=sqrt{x-|x-2|}`

`x>=2=>|x-2|=x-2`

`=>A=sqrt{x-x+2}=sqrt2`

`1<=x<=2=>|x-2|=2x-`

`=>A=\sqrt{x+x-2}=sqrt{2x-2}`

\(a,DKXD:x\ge0\)

\(b,A=\sqrt{x-\sqrt{x^2-4x+4}}\)

\(=\sqrt{x-\sqrt{\left(x-2\right)^2}}\)

\(=\sqrt{x-\left|x-2\right|}\)

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

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

\(=\sqrt{2}\)

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(A=\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\)

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

\(=\dfrac{-4x^2-8x}{\left(x-2\right)\left(x+2\right)}=\dfrac{-4x}{x-2}\)

a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\)      (ĐK : \(\forall x\in R\))

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

     * Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)

     *Nếu x<2   => M=2-x-x-2=-2x

b,Để M=2\(\ne-4\)

     =>M=-2x

    =>-2x=-4

    =>x=2

__________________________________________________________________________________________

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

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

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

     * Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)

    * Nếu x<2  =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)

             VẬY.......

 Tk nha!

Lời giải:

a.

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

ĐKXĐ: \(\left\{\begin{matrix} x\geq 0\\ \sqrt{x}-4\neq 0\\ \sqrt{x}-2\neq 0\\ \sqrt{x}-1\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ x\neq 16\\ x\neq 4\\ x\neq 1\end{matrix}\right.\)

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

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

b.

Với $x$ nguyên, để $A\in\mathbb{Z}$ thì $\sqrt{x}+1\vdots 2(\sqrt{x}-2)}$

$\Rightarrow \sqrt{x}+1\vdots \sqrt{x}-2$
$\Leftrightarrow \sqrt{x}-2+3\vdots \sqrt{x}-2$

$\Leftrightarrow 3\vdots \sqrt{x}-2$

$\Rightarrow \sqrt{x}-2\in\left\{\pm 1;\pm 3\right\}$

$\Rightarrow x\in\left\{1;9;25\right\}$

Thử lại thấy đều thỏa mãn.