`~rai~`

\(b)\sqrt{x^2+4x+5}\\ĐKXĐ:x^2+4x+5\ge 0\\\Leftrightarrow (x^2+4x+4)+1\ge 0\\\Leftrightarrow (x+2)^2\ge -1\quad\text{(luôn đúng vì }(x+2)^2\ge 0\quad\forall x\in\m ...

`~rai~`

\(\sqrt{2x^2+2}=3x-1\quad\left(ĐK:x\ge \dfrac{1}{3}\right)\\\Leftrightarrow \left(\sqrt{2x^2+2}\right)^2=(3x-1)^2\\\Leftrightarrow 2x^2+2=9x^2-6x+1\\\Leftrightarrow 7x^2-6x-1=0\\\Lef ...

`~rai~`

$\dfrac{\sqrt{x^2-6x+9}}{x-3}\quad(x<3)\\=\dfrac{\sqrt{(x-3)^2}}{x-3}\\=\dfrac{|x-3|}{x-3}\\=\dfrac{3-x}{x-3}\quad\text{(vì }x<3)\\=\dfrac{-(x-3)}{x-3}\\=-1.$

`~rai~`

\(\sqrt{x-\sqrt{2x-1}}\\ĐKXĐ:\begin{cases}2x-1\ge 0\\x-\sqrt{2x-1}\ge 0\end{cases}\\\Leftrightarrow \begin{cases}2x\ge 1\\x\ge \sqrt{2x-1}\end{cases}\\\Leftrightarrow \begin{cases}x\ ...

`~rai~`

\(A=\dfrac{x-6\sqrt{x}+9}{x-9}\quad(ĐK:x\ge 0;x\ne 9)\\\text{Ta có:}|A|=-A\\\Leftrightarrow A\le 0\\\Leftrightarrow \dfrac{x-6\sqrt{x}+9}{x-9}\le 0\\\Leftrightarrow \dfrac{(\s ...

`~rai~`

\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\\=\sqrt{3+2.\sqrt{3.2}+2}-\sqrt{3-2\sqrt{3.2}+2}\\=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\\=|\sqrt{3} ...

`~rai~`

\(1.\\d)y=\dfrac{1}{\sqrt{x-4x+4}}\\ĐKXĐ:\begin{cases}x^2-4x+4\ge 0\\x^2-4x+4\ne 0\end{cases}\\\Leftrightarrow x^2-4x+4>0\\\Leftrightarrow (x-2)^2>0\\\Leftrightarrow x-2\ne 0\q ...

`~rai~`

\(c)\dfrac{x}{y}.\sqrt{\dfrac{x^2}{y^4}}\quad(x>0;y\ne 0)\\=\dfrac{x}{y}.\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\\=\dfrac{x}{y}.\dfrac{|x|}{\sqrt{(y^2)^2}}\\=\dfrac{x}{y}.\dfrac{x}{|y^2|}\ ...

`~rai~`

\((\sqrt{x}+1)^2=6\sqrt{x}-3-\sqrt{x-4}\quad(x\ge 4)\\\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\\\Leftrightarrow x+2\sqrt{x}+1-6\sqrt{x}+3+\sqrt{x-4}=0\\\Leftrightarrow x- ...

`~rai~`

\(x^2+(2m-1)x-2m=0\quad(1)\\\text{Phương trình (1) có 2 nghiệm}\Leftrightarrow \begin{cases}a\ne 0\\\Delta\ge0\end{cases}\\\Leftrightarrow \begin{cases}1\ne 0\quad\text ...