Bài 1.
Bạn ghi nhớ để $\sqrt{a}$ xác định thì $a\geq 0$
a. ĐKXĐ:
$-3x+2\geq 0\Leftrightarrow x\leq \frac{2}{3}$
b.
\(\left\{\begin{matrix} 2x+3\neq 0\\ \frac{4}{2x+3}\geq 0\end{matrix}\right.\Leftrightarrow 2x+3>0\Leftrightarrow x>\frac{-3}{2}\)
c.
\(\left\{\begin{matrix} x^2\neq 0\\ \frac{2}{x^2}>0\end{matrix}\right.\Leftrightarrow x\ne 0\)
d.
$x(x+2)\geq 0$
$\Leftrightarrow x>0$ hoặc $x\leq -2$
e.
$9x^2-6x+1\geq 0$
$\Leftrightarrow (3x-1)^2\geq 0\Leftrightarrow x\in\mathbb{R}$
f.
\(\left\{\begin{matrix} 2-x\neq 0\\ \frac{2x-1}{2-x}\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 2\\ 2>x\geq \frac{1}{2}\end{matrix}\right.\Leftrightarrow 2> x\geq \frac{1}{2}\)
g.
$5x^2-3x-8\geq 0$
$\Leftrightarrow (x+1)(5x-8)\geq 0$
$\Leftrightarrow x\geq \frac{8}{5}$ hoặc $x\leq -1$
h.
$5x^2+4x+7\geq 0$
$\Leftrightarrow (2x)^2+(x+2)^2+3\geq 0$
$\Leftrightarrow x\in\mathbb{R}$
Bài 2:
a. $2\sqrt{x^2}=2|x|=-2x$
b. $\frac{1}{2}\sqrt{x^{10}}=\frac{1}{2}|x^5|=-\frac{1}{2}x^5$
c. $\sqrt{(x-5)^2}=|x-5|=5-x$
d. $\sqrt{(x-10)^{10}}=|(x-10)^5|=(10-x)^5$
e. $x-4+\sqrt{x^2-8x+16}=x-4+\sqrt{(x-4)^2}$
$=x-4+|x-4|=x-4+(4-x)=0$
f. $\sqrt{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})^2}$
$=\sqrt{(x-y)^2}=|x-y|=y-x$
g. $\frac{3-\sqrt{x}}{x-9}=\frac{-(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{-1}{\sqrt{x}+3}$
h. $\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}=\frac{(\sqrt{x}-2)(\sqrt{x}-3)}{\sqrt{x}-3}=\sqrt{x}-2$
i. $\sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{(\sqrt{2}+1)^2}=\sqrt{2}+1$
k. $\sqrt{9-4\sqrt{5}}=\sqrt{9-2\sqrt{20}}=\sqrt{4+5-2\sqrt{4.5}}$
$=\sqrt{(\sqrt{5}-\sqrt{4})^2}=\sqrt{5}-\sqrt{4}=\sqrt{5}-2$
Bài 3.
a. $\sqrt{x^2}=8$
$\Leftrightarrow |x|=8$
$\Leftrightarrow x=\pm 8$
b. $\sqrt{(x-5)^2}=|-5|$
$\Leftrightarrow |x-5|=5$
$\Leftrightarrow x-5=\pm 5$
$\Rightarrow x=0$ hoặc $x=10$
c. $\sqrt{1-12x+36x^2}=5$
$\Leftrightarrow \sqrt{(6x-1)^2}=5$
$\Leftrightarrow |6x-1|=5$
$\Rightarrow 6x-1=\pm 5$
$\Rightarrow x=1$ hoặc $x=\frac{-2}{3}$
d.
$\sqrt{(x-3)^2}=3-x$
$\Leftrightarrow |x-3|=3-x$
$\Leftrightarrow 3-x\geq 0$
$\Leftrightarrow x\leq 3$
Bài 3.
e.
$\sqrt{25-20x+4x^2}+2x=5$
$\Leftrightarrow \sqrt{(2x-5)^2}=5-2x$
$\Leftrightarrow |2x-5|=5-2x$
$\Leftrightarrow 5-2x\geq 0$
$\Leftrightarrow x\leq \frac{5}{2}$
g.
$\sqrt{x+2\sqrt{x-1}}=2$
$\Leftrightarrow \sqrt{(x-1)+2\sqrt{x-1}+1}=2$
$\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2$
$\Leftrightarrow |\sqrt{x-1}+1|=2$
$\Rightarrow \sqrt{x-1}+1=\pm 2$
TH $\sqrt{x-1}+1=-2$ loại do $\sqrt{x-1}+1\geq 1$.
$\Rightarrow \sqrt{x-1}+1=2$
$\Leftrightarrow x=2$
Bài 4:
a.
$9+4\sqrt{5}=9+2\sqrt{20}=4+5+2\sqrt{4}.\sqrt{5}=(\sqrt{4}+\sqrt{5})^2=(2+\sqrt{5})^2$
đpcm
b.
$\sqrt{9+4\sqrt{5}}-\sqrt{5}=\sqrt{(2+\sqrt{5})^2}-\sqrt{5}$
$=|2+\sqrt{5}|-\sqrt{5}=2+\sqrt{5}-\sqrt{5}=2$
c.
$\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{16+7+2.\sqrt{16}.\sqrt{7}}-\sqrt{7}$
$=\sqrt{(\sqrt{16}+\sqrt{7})^2}-\sqrt{7}=|\sqrt{16}+\sqrt{7}|-\sqrt{7}$
$=\sqrt{16}+\sqrt{7}-\sqrt{7}=4$
d. Câu này không đúng.
$\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}$
$=\sqrt{(a-2)+4\sqrt{a-2}+4}+\sqrt{(a-2)-4\sqrt{a-2}+4}$
$=\sqrt{(\sqrt{a-2}+2)^2}+\sqrt{(\sqrt{a-2}-2)^2}$
$=|\sqrt{a-2}+2|+|\sqrt{a-2}-2|$
Nếu $a\geq 4$ thì:
$\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=\sqrt{a-2}+2+\sqrt{a-2}-2=2\sqrt{a-2}$
Nếu $2\leq a< 4$ thì:
$\sqrt{a+4\sqrt{a-2}+2}+\sqrt{a-4\sqrt{a-2}+2}=\sqrt{a-2}+2+2-\sqrt{a-2}=4$
