Bài 2:
a.
\(\sqrt{8}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)
\(\sqrt{65}-1> \sqrt{64}-1=8-1=7\)
\(\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1\)
b.
\(13-2\sqrt{3}=9+4-2\sqrt{3}=9+2\sqrt{4}-2\sqrt{3}>9\)
\(6\sqrt{2}= 6\sqrt{\frac{8}{4}}< 6\sqrt{\frac{9}{4}}=6.\frac{3}{2}=9\)
$\Rightarrow 13-2\sqrt{3}> 6\sqrt{2}$
$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$
Bài 3:
a. ĐKXĐ: $9-x^2\geq 0$
$\Leftrightarrow (3-x)(3+x)\geq 0$
$\Leftrightarrow 3\geq x\geq -3$
b. ĐKXĐ: $-4x^2+4x-1\geq 0$
$\Leftrightarrow 4x^2-4x+1\leq 0$
$\Leftrightarrow (2x-1)^2\leq 0$
Mà $(2x-1)^2\geq 0$ với mọi $x\in\mathbb{R}$
$\Rightarrow (2x-1)^2=0$
$\Leftrightarrow x=\frac{1}{2}$
c.
ĐKXĐ: $x^2+x-2>0$
$\Leftrightarrow (x-1)(x+2)>0$
$\Leftrightarrow x>1$ hoặc $x< -2$
Bài 3:
a) ĐKXĐ: \(-3\le x\le3\)
b) ĐKXĐ: \(x=\dfrac{1}{2}\)
c) ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
